Real Analysis

1. The Real Number System

1.1 Field Axioms

The real numbers $\mathbb{R}$ form a complete ordered field. The field axioms guarantee closure Under addition, subtraction, multiplication, and division (by non-zero elements), together with the Usual commutative, associative, and distributive laws.

1.2 Order and the Completeness Axiom

The order relation $\leq$ on $\mathbb{R}$ satisfies:

Reflexivity: $a \leq a$
Antisymmetry: $a \leq b$ and $b \leq a$ implies $a = b$
Transitivity: $a \leq b$ and $b \leq c$ implies $a \leq c$
Totality: for all $a, b$ Either $a \leq b$ or $b \leq a$
Compatibility with addition: $a \leq b$ implies $a + c \leq b + c$
Compatibility with multiplication: $a \leq b$ and $0 \leq c$ implies $ac \leq bc$

The completeness axiom (also called the least upper bound property) is what distinguishes $\mathbb{R}$ from $\mathbb{Q}$ :

Axiom (Completeness). Every non-empty subset of $\mathbb{R}$ that is bounded above has a least Upper bound (supremum) in $\mathbb{R}$ .

1.3 Supremum and Infimum

Let $S \subseteq \mathbb{R}$ be a non-empty set that is bounded above.

Definition. The supremum (or least upper bound) of $S$ Denoted $\sup(S)$ Is the real number $u$ satisfying:

$u$ is an upper bound: $s \leq u$ for all $s \in S$ .
$u$ is the least upper bound: if $v$ is any upper bound of $S$ Then $u \leq v$ .

Similarly, the infimum (or greatest lower bound), $\inf(S)$ Is the greatest number $l$ such that $l \leq s$ for all $s \in S$ .

Proposition 1.1. $\sup(S)$ exists if and only if $S$ is non-empty and bounded above.

Proposition 1.2 (Approximation Property). If $u = \sup(S)$ Then for every $\varepsilon > 0$ There Exists $s \in S$ such that $u - \varepsilon \lt s \leq u$ .

Proof. If no such $s$ existed, then $u - \varepsilon$ would be an upper bound of $S$ strictly less Than $u$ Contradicting the definition of $\sup(S)$ . $\blacksquare$

Example. Let $S = \{x \in \mathbb{R} : x^2 \lt 2\}$ . Then $\sup(S) = \sqrt{2}$ . Note that $\sqrt{2} \notin \mathbb{Q}$ , so $\mathbb{Q}$ does not satisfy the completeness axiom.

1.4 Archimedean Property

Theorem 1.1 (Archimedean Property). For every $x \in \mathbb{R}$ There exists $n \in \mathbb{N}$ Such that $n \gt x$ .

Proof. Suppose, for contradiction, that $\mathbb{N}$ is bounded above. By the completeness axiom, $s = \sup(\mathbb{N})$ exists in $\mathbb{R}$ . Then $s - 1$ is not an upper bound for $\mathbb{N}$ So there exists $n \in \mathbb{N}$ with $n \gt s - 1$ I.e., $n + 1 \gt s$ . But $n + 1 \in \mathbb{N}$ Contradicting that $s$ is an upper bound. $\blacksquare$

Corollary 1.2. For every $\varepsilon > 0$ There exists $n \in \mathbb{N}$ such that $1/n \lt \varepsilon$ .

Proof. By the Archimedean property, choose $n \in \mathbb{N}$ with $n \gt 1/\varepsilon$ . Then $1/n \lt \varepsilon$ . $\blacksquare$

Corollary 1.3 (Density of $\mathbb{Q}$ ). Between any two distinct real numbers $a \lt b$ There Exists a rational number $q \in \mathbb{Q}$ with $a \lt q \lt b$ .

Proof. Since $b - a > 0$ By Corollary 1.2 there exists $n \in \mathbb{N}$ with $1/n \lt b - a$ So $1 \lt n(b - a) = nb - na$ . Let $m = \lfloor na \rfloor + 1 \in \mathbb{Z}$ . Then $m - 1 \leq na \lt m$ Giving $m \leq na + 1 \lt na + n(b - a) = nb$ . Hence $a \lt m/n \lt b$ And $m/n \in \mathbb{Q}$ . $\blacksquare$

1.5 Properties of Supremum and Infimum

Proposition 1.4. If $A$ and $B$ are non-empty bounded subsets of $\mathbb{R}$ Then $\sup(A + B) = \sup(A) + \sup(B)$ Where $A + B = \{a + b : a \in A, b \in B\}$ .

Proof. For all $a \in A$ and $b \in B$ : $a \leq \sup(A)$ and $b \leq \sup(B)$ So $a + b \leq \sup(A) + \sup(B)$ . Thus $\sup(A) + \sup(B)$ is an upper bound for $A + B$ So $\sup(A + B) \leq \sup(A) + \sup(B)$ .

For the reverse inequality, let $\varepsilon > 0$ . By the approximation property, there exist $a \in A$ And $b \in B$ with $a > \sup(A) - \varepsilon/2$ and $b > \sup(B) - \varepsilon/2$ . Then $a + b > \sup(A) + \sup(B) - \varepsilon$ So $\sup(A + B) \geq \sup(A) + \sup(B) - \varepsilon$ . Since $\varepsilon > 0$ is arbitrary, $\sup(A + B) \geq \sup(A) + \sup(B)$ . $\blacksquare$

Proposition 1.5. For any non-empty bounded set $S \subseteq \mathbb{R}$ , $\inf(S) = -\sup(-S)$ Where $-S = \{-s : s \in S\}$ .

Proof. Let $u = \sup(-S)$ . Then $-s \leq u$ for all $s \in S$ So $s \geq -u$ for all $s \in S$ Meaning $-u$ is a lower bound for $S$ . If $v$ is any lower bound for $S$ Then $-v$ is an upper bound For $-S$ So $u \leq -v$ I.e., $-u \geq v$ . Hence $-u = \inf(S)$ . $\blacksquare$

Worked Example: Find $\sup$ and $\inf$ of $S = \{(-1)^n + 1/n : n \in \mathbb{N}\}$

Solution. The first few terms are $0, 3/2, -2/3, 5/4, -4/5, 7/6, \ldots$ .

For even $n = 2k$ : $(-1)^{2k} + 1/(2k) = 1 + 1/(2k)$ Which decreases toward $1$ from above. For odd $n = 2k-1$ : $(-1)^{2k-1} + 1/(2k-1) = -1 + 1/(2k-1)$ Which increases toward $-1$ from below.

The even terms form the sequence $3/2, 5/4, 7/6, \ldots$ with limit $1$ So $\sup(S) = 3/2$ (the first even term). The odd terms form $0, -2/3, -4/5, \ldots$ with limit $-1$ And since $0$ Is an odd-indexed term, $\inf(S) = -1$ (approached but not attained). $\blacksquare$

1.6 Construction of $\mathbb{R}$ via Dedekind Cuts

Remark. The following outline shows how $\mathbb{R}$ can be constructed from $\mathbb{Q}$ Making The completeness axiom a theorem rather than an axiom.

Definition (Dedekind Cut). A Dedekind cut is a subset $\alpha \subseteq \mathbb{Q}$ satisfying:

$\alpha \neq \emptyset$ and $\alpha \neq \mathbb{Q}$
If $p \in \alpha$ and $q \lt p$ (with $q \in \mathbb{Q}$ ), then $q \in \alpha$ (downward closure)
$\alpha$ has no greatest element: for every $p \in \alpha$ There exists $q \in \alpha$ with $p \lt q$

Definition. The set of real numbers $\mathbb{R}$ is defined as the set of all Dedekind cuts.

The order, addition, and multiplication are defined as follows:

Order: $\alpha \lt \beta$ if and only if $\alpha \subsetneq \beta$
Addition: $\alpha + \beta = \{p + q : p \in \alpha, q \in \beta\}$
Multiplication: For $\alpha, \beta \geq 0^*$ : $\alpha \cdot \beta = \{p \cdot q : p \in \alpha, q \in \beta, p \geq 0, q \geq 0\} \cup \{r \in \mathbb{Q} : r \lt 0\}$

Here $0^* = \{q \in \mathbb{Q} : q \lt 0\}$ represents the real number $0$ .

Theorem. With these definitions, $\mathbb{R}$ is a complete ordered field, and $\mathbb{Q}$ embeds Into $\mathbb{R}$ via $q \mapsto \{r \in \mathbb{Q} : r \lt q\}$ .

Proof (sketch). Verifying the field axioms and order axioms is lengthy but straightforward. The key Step is the completeness axiom: if $\mathcal{A}$ is a non-empty set of Dedekind cuts bounded above, Then $\alpha = \bigcup_{\beta \in \mathcal{A}} \beta$ is itself a Dedekind cut and $\alpha = \sup(\mathcal{A})$ . $\blacksquare$

1.7 Equivalences of Completeness

The completeness axiom can be formulated in several equivalent ways. Each implies the others:

Least Upper Bound Property: Every non-empty set bounded above has a supremum.
Monotone Convergence Theorem: Every bounded monotone sequence converges.
Nested Interval Property: Every nested sequence of closed intervals $I_1 \supseteq I_2 \supseteq \cdots$ with $\mathrm{length}(I_n) \to 0$ has exactly one point in $\bigcap I_n$ .
Bolzano-Weierstrass Property: Every bounded sequence has a convergent subsequence.
Cauchy Completeness: Every Cauchy sequence converges.

Proposition 1.6. In any ordered field, (1) $\iff$ (2) $\iff$ (3) $\iff$ (4) $\iff$ (5).

Proof (outline). We have shown $(1) \Rightarrow (2)$ (MCT in Section 2.2), $(2) \Rightarrow (4)$ (via the bisection argument in Bolzano-Weierstrass), $(4) \Rightarrow (5)$ (Cauchy completeness proof In Section 2.3), and $(5) \Rightarrow (1)$ can be shown by constructing a Cauchy sequence converging To $\sup S$ from the approximation property. The equivalence $(1) \Rightarrow (3)$ follows from the Nested interval argument, and $(3) \Rightarrow (1)$ follows by constructing nested intervals that Shrink to $\sup S$ . $\blacksquare$

Remark. The field $\mathbb{Q}$ satisfies none of these properties, which is why it must be Extended to $\mathbb{R}$ for analysis.

:::caution Common Pitfall The completeness axiom is often misstated as “every bounded set has a supremum.” The set must be Non-empty. Also, completeness does not say every set has a maximum; $\sup(S)$ need not belong to $S$ . For example, $\sup\{1/n : n \in \mathbb{N}\} = 1$ Which belongs to the set, but $\sup(0, 1) = 1$ Which does not belong to $(0, 1)$ . :::

2. Sequences and Limits

2.1 Convergence

A sequence $(a_n)_{n=1}^{\infty}$ in $\mathbb{R}$ converges to a limit $L \in \mathbb{R}$ if for Every $\varepsilon > 0$ There exists $N \in \mathbb{N}$ such that

$|a_n - L| \lt \varepsilon \quad \mathrm{for\ all\ } n \geq N$

We write $a_n \to L$ or $\lim_{n \to \infty} a_n = L$ . A sequence that does not converge is said to diverge.

Proposition 2.1 (Uniqueness of Limits). If $(a_n)$ converges, its limit is unique.

Proof. Suppose $a_n \to L$ and $a_n \to M$ with $L \neq M$ . Let $\varepsilon = |L - M|/2 > 0$ . There Exists $N_1$ such that $|a_n - L| \lt \varepsilon$ for $n \geq N_1$ And $N_2$ such that $|a_n - M| \lt \varepsilon$ for $n \geq N_2$ . For $n \geq \max(N_1, N_2)$ :

$|L - M| \leq |a_n - L| + |a_n - M| \lt 2\varepsilon = |L - M|$

A contradiction. $\blacksquare$

Proposition 2.2. Every convergent sequence is bounded.

Proof. Let $a_n \to L$ . Taking $\varepsilon = 1$ There exists $N$ such that $|a_n - L| \lt 1$ for All $n \geq N$ . Then $|a_n| \leq |L| + 1$ for $n \geq N$ . Let $M = \max\{|a_1|, |a_2|, \ldots, |a_{N-1}|, |L| + 1\}$ . Then $|a_n| \leq M$ for all $n$ . $\blacksquare$

2.2 Convergence Theorems

Theorem 2.1 (Algebra of Limits). If $a_n \to L$ and $b_n \to M$ Then:

$a_n + b_n \to L + M$
$a_n b_n \to LM$
$a_n / b_n \to L/M$ (provided $M \neq 0$ and $b_n \neq 0$ for all $n$ )

Theorem 2.2 (Squeeze Theorem). If $a_n \leq b_n \leq c_n$ for all $n$ and $a_n \to L$ $c_n \to L$ Then $b_n \to L$ .

Theorem 2.3 (Monotone Convergence Theorem). Every bounded monotone sequence in $\mathbb{R}$ converges. Specifically:

Every bounded increasing sequence converges to its supremum.
Every bounded decreasing sequence converges to its infimum.

Proof. Let $(a_n)$ be bounded and increasing. By the completeness axiom, $s = \sup\{a_n : n \in \mathbb{N}\}$ exists. Let $\varepsilon > 0$ . By the approximation property, There exists $N$ such that $s - \varepsilon \lt a_N \leq s$ . Since $(a_n)$ is increasing, $a_n \geq a_N > s - \varepsilon$ for all $n \geq N$ . Also $a_n \leq s$ for all $n$ . Hence $|a_n - s| \lt \varepsilon$ for all $n \geq N$ . $\blacksquare$

2.3 Cauchy Sequences

A sequence $(a_n)$ is a Cauchy sequence if for every $\varepsilon > 0$ There exists $N \in \mathbb{N}$ such that

$|a_n - a_m| \lt \varepsilon \quad \mathrm{for\ all\ } m, n \geq N$

Theorem 2.4. Every convergent sequence is Cauchy.

Proof. Let $a_n \to L$ . Given $\varepsilon > 0$ Choose $N$ such that $|a_n - L| \lt \varepsilon/2$ For all $n \geq N$ . Then for $m, n \geq N$ : $|a_n - a_m| \leq |a_n - L| + |a_m - L| \lt \varepsilon$ . $\blacksquare$

Theorem 2.5 (Cauchy Completeness of $\mathbb{R}$ ). Every Cauchy sequence in $\mathbb{R}$ converges.

Proof. Let $(a_n)$ be Cauchy. First, $(a_n)$ is bounded: choose $N$ with $|a_n - a_m| \lt 1$ for $m, n \geq N$ . Then $|a_n| \leq |a_N| + 1$ for $n \geq N$ . By the Bolzano-Weierstrass theorem (Theorem 2.6 below), $(a_n)$ has a convergent subsequence $(a_{n_k}) \to L$ . We show $a_n \to L$ .

Given $\varepsilon > 0$ Choose $N_1$ so that $|a_n - a_m| \lt \varepsilon/2$ for $m, n \geq N_1$ And $K$ so that $|a_{n_k} - L| \lt \varepsilon/2$ for $k \geq K$ . For $n \geq N_1$ Choose $k \geq K$ with $n_k \geq N_1$ (possible since $n_k \to \infty$ ). Then

$|a_n - L| \leq |a_n - a_{n_k}| + |a_{n_k} - L| \lt \varepsilon/2 + \varepsilon/2 = \varepsilon$

$\blacksquare$

2.4 Subsequences

A subsequence of $(a_n)$ is a sequence $(a_{n_k})_{k=1}^{\infty}$ where $n_1 \lt n_2 \lt n_3 \lt \cdots$ .

Proposition 2.3. If $a_n \to L$ Then every subsequence $(a_{n_k}) \to L$ .

Proposition 2.4. If $(a_n)$ has two subsequences converging to different limits, then $(a_n)$ diverges.

2.5 The Bolzano-Weierstrass Theorem

Theorem 2.6 (Bolzano-Weierstrass). Every bounded sequence in $\mathbb{R}$ has a convergent subsequence.

Proof. Let $(a_n)$ be bounded, so $a_n \in [A, B]$ for all $n$ . Set $I_0 = [A, B]$ . Bisect $I_0$ into $[A, (A+B)/2]$ and $[(A+B)/2, B]$ . At least one contains infinitely many terms of $(a_n)$ ; call it $I_1$ . Having constructed $I_k = [l_k, r_k]$ Bisect it and select $I_{k+1}$ as the half containing Infinitely many terms of $(a_n)$ .

This produces a nested sequence of closed intervals $I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots$ With $\mathrm{length}(I_k) = (B - A)/2^k \to 0$ . By the Nested Interval Property (which follows From completeness), $\bigcap_{k=0}^{\infty} I_k = \{c\}$ for some $c \in [A, B]$ .

Construct the subsequence inductively: pick $n_1$ with $a_{n_1} \in I_1$ . Having chosen $n_1 \lt n_2 \lt \cdots \lt n_{k-1}$ Pick $n_k > n_{k-1}$ with $a_{n_k} \in I_k$ (possible since $I_k$ contains infinitely many terms). Then $a_{n_k} \in I_k$ for all $k$ So $|a_{n_k} - c| \leq \mathrm{length}(I_k) \to 0$ . Hence $a_{n_k} \to c$ . $\blacksquare$

2.6 Limit Superior and Limit Inferior

Let $(a_n)$ be a bounded sequence. Define:

$\limsup_{n \to \infty} a_n = \inf_{n \geq 1} \sup_{k \geq n} a_k, \qquad \liminf_{n \to \infty} a_n = \sup_{n \geq 1} \inf_{k \geq n} a_k$

Proposition 2.5. For every bounded sequence $(a_n)$ : $\liminf_{n \to \infty} a_n \leq \limsup_{n \to \infty} a_n$

Proof. For any $n$ , $\inf_{k \geq n} a_k \leq a_n \leq \sup_{k \geq n} a_n$ . Taking supremum over $n$ on the left: $\liminf a_n \leq \sup_{k \geq n} a_k$ for every $n$ . Taking infimum over $n$ on The right gives $\liminf a_n \leq \limsup a_n$ . $\blacksquare$

Proposition 2.6. $(a_n)$ converges if and only if $\liminf a_n = \limsup a_n$ In which case the Common value equals $\lim a_n$ .

Proof. If $a_n \to L$ Then for every $\varepsilon > 0$ There exists $N$ such that $L - \varepsilon \lt a_n \lt L + \varepsilon$ for $n \geq N$ . Hence $\sup_{k \geq n} a_k \leq L + \varepsilon$ For $n \geq N$ So $\limsup a_n \leq L + \varepsilon$ . Since $\varepsilon > 0$ is arbitrary, $\limsup a_n \leq L$ . Similarly $\liminf a_n \geq L$ . Combined with Proposition 2.5, $\liminf a_n = \limsup a_n = L$ .

Conversely, if $\liminf a_n = \limsup a_n = L$ Then for every $\varepsilon > 0$ There exists $N_1$ With $\sup_{k \geq n} a_k \lt L + \varepsilon$ for $n \geq N_1$ And $N_2$ with $\inf_{k \geq n} a_k > L - \varepsilon$ for $n \geq N_2$ . For $n \geq \max(N_1, N_2)$ : $L - \varepsilon \lt a_n \lt L + \varepsilon$ So $a_n \to L$ . $\blacksquare$

Proposition 2.7. $\limsup a_n$ is the largest subsequential limit of $(a_n)$ And $\liminf a_n$ Is the smallest.

Proof. Let $L^* = \limsup a_n = \inf_n \sup_{k \geq n} a_k$ . Define $s_n = \sup_{k \geq n} a_k$ . Then $(s_n)$ is decreasing and $s_n \to L^*$ . For each $n$ Choose $k_n \geq n$ with $a_{k_n} > s_n - 1/n$ . Then $a_{k_n} \to L^*$ (by squeeze), producing a subsequence converging to $L^*$ .

If $L > L^*$ were a subsequential limit, choose a subsequence $a_{n_j} \to L$ . For large $j$ : $a_{n_j} > (L + L^*)/2 > L^*$ . But $a_{n_j} \leq s_{n_j}$ for all $j$ And $s_{n_j} \to L^*$ So $a_{n_j} \leq s_{n_j} \lt (L + L^*)/2$ for large $j$ A contradiction. $\blacksquare$

Proposition 2.8 (Algebra of $\limsup$ / $\liminf$ ). If $(a_n)$ and $(b_n)$ are bounded sequences:

$\limsup(a_n + b_n) \leq \limsup a_n + \limsup b_n$
$\liminf(a_n + b_n) \geq \liminf a_n + \liminf b_n$
If $a_n \geq 0$ and $b_n \geq 0$ : $\limsup(a_n b_n) \leq (\limsup a_n)(\limsup b_n)$

Remark. Equality in (1) does not hold . For example, $a_n = (-1)^n$ and $b_n = (-1)^{n+1}$ Give $a_n + b_n = 0$ So $\limsup(a_n + b_n) = 0 \lt 1 + 1 = \limsup a_n + \limsup b_n$ .

Proposition 2.9. A sequence $(a_n)$ is convergent if and only if it is Cauchy, if and only if $\limsup a_n = \liminf a_n$ .

Worked Example: Compute $\limsup$ and $\liminf$ of $a_n = (-1)^n \cdot \frac{n}{n+1}$

Solution. The sequence is $-1/2, 2/3, -3/4, 4/5, -5/6, \ldots$

The even subsequence is $a_{2k} = \frac{2k}{2k+1} \to 1$ . The odd subsequence is $a_{2k-1} = -\frac{2k-1}{2k} \to -1$ .

No subsequence can have a limit greater than $1$ (since $a_n \leq n/(n+1) \lt 1$ for even $n$ And $a_n \lt 0$ for odd $n$ ). Similarly, no subsequence can have a limit less than $-1$ .

Therefore $\limsup_{n \to \infty} a_n = 1$ and $\liminf_{n \to \infty} a_n = -1$ . Since $\limsup \neq \liminf$ The sequence diverges. $\blacksquare$

2.7 Worked Examples

Problem. Prove that $\lim_{n \to \infty} \frac{n}{n+1} = 1$ .

Solution. Let $\varepsilon > 0$ . We need $\left|\frac{n}{n+1} - 1\right| \lt \varepsilon$ I.e., $\frac{1}{n+1} \lt \varepsilon$ I.e., $n > \frac{1}{\varepsilon} - 1$ . Choose $N = \lceil \frac{1}{\varepsilon} \rceil$ . Then for $n \geq N$ : $n \geq \frac{1}{\varepsilon}$ So $n+1 > \frac{1}{\varepsilon}$ So $\frac{1}{n+1} \lt \varepsilon$ . $\blacksquare$

Worked Example: $\varepsilon$-$N$ proof for $\lim_{n \to \infty} \frac{3n + 1}{n + 2} = 3$

Solution. Let $\varepsilon > 0$ . We compute:

$\left|\frac{3n+1}{n+2} - 3\right| = \left|\frac{3n+1 - 3(n+2)}{n+2}\right| = \left|\frac{-5}{n+2}\right| = \frac{5}{n+2}$

We need $\frac{5}{n+2} \lt \varepsilon$ I.e., $n + 2 > 5/\varepsilon$ I.e., $n > 5/\varepsilon - 2$ . Choose $N = \lceil 5/\varepsilon \rceil$ . Then for $n \geq N$ :

$\left|\frac{3n+1}{n+2} - 3\right| = \frac{5}{n+2} \leq \frac{5}{N+2} \leq \frac{5}{5/\varepsilon} = \varepsilon$

$\blacksquare$

Worked Example: Show $(a_n)$ with $a_1 = \sqrt{2}$, $a_{n+1} = \sqrt{2 + a_n}$ converges

Solution. Step 1: $(a_n)$ is bounded above by $2$ . By induction: $a_1 = \sqrt{2} \leq 2$ . If $a_n \leq 2$ Then $a_{n+1} = \sqrt{2 + a_n} \leq \sqrt{2 + 2} = 2$ .

Step 2: $(a_n)$ is increasing. We have $a_1 = \sqrt{2} \approx 1.414$ and $a_2 = \sqrt{2 + \sqrt{2}} \approx 1.848$ . Assume $a_n \leq a_{n+1}$ . Then $a_{n+1} = \sqrt{2 + a_n} \leq \sqrt{2 + a_{n+1}} = a_{n+2}$ .

Step 3: By the Monotone Convergence Theorem, $(a_n)$ converges. Let $L = \lim a_n$ . Taking limits In $a_{n+1} = \sqrt{2 + a_n}$ : $L = \sqrt{2 + L}$ So $L^2 = 2 + L$ Giving $L^2 - L - 2 = 0$ So $(L-2)(L+1) = 0$ . Since $a_n \geq \sqrt{2} > 0$ for all $n$ , $L \geq 0$ So $L = 2$ . $\blacksquare$

:::caution Common Pitfall When computing $\limsup$ and $\liminf$ Do not confuse them with $\sup$ and $\inf$ of the range $\{a_n : n \in \mathbb{N}\}$ . The $\limsup$ depends on the tail behavior of the sequence. For Example, $a_n = (-1)^n$ has $\limsup = 1$ and $\liminf = -1$ But $\sup\{a_n\} = 1$ and $\inf\{a_n\} = -1$ happen to agree in this case. However, for $a_n = 1/n$ , $\sup = 1$ but $\limsup = 0$ . :::

3. Series

3.1 Definitions and Convergence

A series $\sum_{n=1}^{\infty} a_n$ converges if the sequence of partial sums $S_N = \sum_{n=1}^{N} a_n$ Converges. The limit is the sum of the series.

If $a_n \geq 0$ for all $n$ The series of partial sums is increasing, so by the monotone convergence Theorem, $\sum a_n$ converges if and only if $(S_N)$ is bounded above.

3.2 Convergence Tests

Theorem 3.1 (Comparison Test). If $0 \leq a_n \leq b_n$ for all $n$ Then:

If $\sum b_n$ converges, then $\sum a_n$ converges.
If $\sum a_n$ diverges, then $\sum b_n$ diverges.

Theorem 3.2 (Limit Comparison Test). If $a_n > 0$ , $b_n > 0$ And $\lim_{n \to \infty} a_n / b_n = L$ where $0 \lt L \lt \infty$ Then $\sum a_n$ converges if and only if $\sum b_n$ converges.

Theorem 3.3 (Ratio Test). If $\lim_{n \to \infty} |a_{n+1}/a_n| = L$ Then:

If $L \lt 1$ , $\sum a_n$ converges absolutely.
If $L > 1$ , $\sum a_n$ diverges.
If $L = 1$ The test is inconclusive.

Theorem 3.4 (Root Test). If $\limsup_{n \to \infty} \sqrt[n]{|a_n|} = L$ Then:

If $L \lt 1$ , $\sum a_n$ converges absolutely.
If $L > 1$ , $\sum a_n$ diverges.
If $L = 1$ The test is inconclusive.

Proof. If $L \lt 1$ Choose $r$ with $L \lt r \lt 1$ . By definition of $\limsup$ There exists $N$ such that $\sqrt[n]{|a_n|} \lt r$ for all $n \geq N$ I.e., $|a_n| \lt r^n$ . Since $\sum r^n$ converges (geometric series with $r \lt 1$ ), the comparison test gives absolute convergence.

If $L > 1$ Then for infinitely many $n$ : $\sqrt[n]{|a_n|} > 1$ So $|a_n| > 1$ . Hence $a_n \not\to 0$ And the series diverges. $\blacksquare$

Theorem 3.5 (Integral Test). If $f : [1, \infty) \to [0, \infty)$ is positive, continuous, and Decreasing, then $\sum_{n=1}^{\infty} f(n)$ converges if and only if $\int_1^{\infty} f(x)\, dx$ converges.

Proof. Since $f$ is decreasing, for $k \leq x \leq k+1$ : $f(k+1) \leq f(x) \leq f(k)$ . Integrating:

$f(k+1) = \int_k^{k+1} f(k+1)\, dx \leq \int_k^{k+1} f(x)\, dx \leq \int_k^{k+1} f(k)\, dx = f(k)$

Summing from $k = 1$ to $N - 1$ :

$\sum_{k=2}^{N} f(k) \leq \int_1^N f(x)\, dx \leq \sum_{k=1}^{N-1} f(k)$

If $\int_1^{\infty} f$ converges, the left inequality shows $\sum f(k)$ is bounded above, hence converges. If $\int_1^{\infty} f$ diverges, the right inequality shows $\sum f(k)$ is unbounded, hence diverges. $\blacksquare$

Theorem 3.6 (Alternating Series Test). If $a_n > 0$ , $a_n$ decreases, and $a_n \to 0$ Then $\sum_{n=1}^{\infty} (-1)^{n+1} a_n$ converges.

Proof. The partial sums of the even-indexed subsequence satisfy $S_{2n} = S_{2n-2} - a_{2n-1} + a_{2n}$ . Since $a_{2n-1} \geq a_{2n}$ We have $S_{2n} \leq S_{2n-2}$ So $(S_{2n})$ is decreasing. Similarly, $(S_{2n+1})$ is increasing. Also $S_{2n+1} = S_{2n} + a_{2n+1} \geq S_{2n}$ . Both sequences are bounded (since $(S_{2n})$ is decreasing and bounded below by $S_1$ And $(S_{2n+1})$ is increasing and bounded Above by $S_2$ ). Hence both converge. Since $a_{2n+1} \to 0$ Their limits coincide. $\blacksquare$

3.3 Absolute and Conditional Convergence

A series $\sum a_n$ converges absolutely if $\sum |a_n|$ converges. It converges conditionally If $\sum a_n$ converges but $\sum |a_n|$ diverges.

Theorem 3.7. If $\sum a_n$ converges absolutely, then $\sum a_n$ converges.

Proof. Since $\sum |a_n|$ converges, the partial sums of $\sum |a_n|$ satisfy the Cauchy criterion. Given $\varepsilon > 0$ There exists $N$ such that for $m > n \geq N$ : $\sum_{k=n+1}^{m} |a_k| \lt \varepsilon$ . Then $\left|\sum_{k=n+1}^{m} a_k\right| \leq \sum_{k=n+1}^{m} |a_k| \lt \varepsilon$ So $\sum a_n$ satisfies The Cauchy criterion and converges. $\blacksquare$

3.4 The Alternating Series Estimation Theorem

Theorem 3.8 (Alternating Series Estimation). If $S = \sum_{n=1}^{\infty} (-1)^{n+1} a_n$ satisfies the Hypotheses of the alternating series test, then the error after $N$ terms satisfies:

$|S - S_N| \leq a_{N+1}$

Proof. We have $S_{2n} \leq S \leq S_{2n+1} = S_{2n} + a_{2n+1}$ and $S_{2n-1} \geq S \geq S_{2n} = S_{2n-1} - a_{2n}$ . In both cases $|S - S_N| \leq a_{N+1}$ . $\blacksquare$

3.5 Cauchy Condensation Test

Theorem 3.8b (Cauchy Condensation Test). If $(a_n)$ is a non-negative, decreasing sequence, then $\sum_{n=1}^{\infty} a_n$ converges if and only if $\sum_{k=0}^{\infty} 2^k a_{2^k}$ converges.

Proof. Group the terms of $\sum a_n$ . For the lower bound, note:

$a_1 + (a_2 + a_3) + (a_4 + a_5 + a_6 + a_7) + \cdots \geq a_1 + 2a_2 + 4a_4 + 8a_8 + \cdots = \sum_{k=0}^{\infty} 2^k a_{2^k}$

Since each group $(a_{2^k} + \cdots + a_{2^{k+1}-1})$ has $2^k$ terms, each $\geq a_{2^k}$ . For the upper Bound:

$a_1 + a_2 + (a_3 + a_4) + (a_5 + a_6 + a_7 + a_8) + \cdots \leq a_1 + a_2 + 2a_4 + 4a_8 + \cdots \leq a_1 + 2\sum_{k=1}^{\infty} 2^{k-1} a_{2^k}$

If $\sum 2^k a_{2^k}$ converges, the upper bound shows $\sum a_n$ converges. If $\sum a_n$ Converges, the lower bound shows $\sum 2^k a_{2^k}$ converges. $\blacksquare$

Corollary. $\sum_{n=1}^{\infty} 1/n^p$ converges if and only if $p > 1$ . Apply the condensation Test: $\sum 2^k \cdot 1/(2^k)^p = \sum 2^{k(1-p)}$ A geometric series with ratio $2^{1-p}$ Which converges iff $1 - p \lt 0$ I.e., $p > 1$ .

3.6 Rearrangement of Series

Theorem 3.9 (Riemann Rearrangement Theorem). If $\sum a_n$ converges conditionally, then for any $L \in \mathbb{R}$ (or $\pm\infty$ ), there exists a rearrangement $\sigma : \mathbb{N} \to \mathbb{N}$ such That $\sum_{n=1}^{\infty} a_{\sigma(n)} = L$ .

Proof (outline). Let $P = \{n : a_n > 0\}$ and $N = \{n : a_n \lt 0\}$ . Since $\sum a_n$ converges Conditionally, both $\sum_{n \in P} a_n = +\infty$ and $\sum_{n \in N} a_n = -\infty$ .

To achieve sum $L \in \mathbb{R}$ : take positive terms in order until the partial sum exceeds $L$ Then take negative terms until it falls below $L$ Then positive terms again, and so on. Since both The positive and negative subseries diverge, this process can always continue. The terms tend to Zero (since the series converges), so the oscillations around $L$ shrink to zero. $\blacksquare$

Remark. By contrast, every rearrangement of an absolutely convergent series converges to the same sum.

3.7 Worked Examples

Problem. Determine whether $\sum_{n=1}^{\infty} \frac{n}{2^n}$ converges.

Solution. Apply the ratio test:

$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{(n+1)/2^{n+1}}{n/2^n} = \lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2} \lt 1$

By the ratio test, the series converges absolutely. $\blacksquare$

Worked Example: Determine convergence of $\sum_{n=1}^{\infty} \frac{1}{n^2 + n}$

Solution. Note that $\frac{1}{n^2 + n} = \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ . This is a Telescoping series. The $N$ -th partial sum is:

$S_N = \sum_{n=1}^{N} \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{N+1}$

Therefore $\lim_{N \to \infty} S_N = 1$ And the series converges to $1$ . $\blacksquare$

Worked Example: Does $\sum_{n=2}^{\infty} \frac{1}{n \ln n}$ converge?

Solution. Apply the integral test with $f(x) = 1/(x \ln x)$ on $[2, \infty)$ . The function is positive, Continuous, and decreasing. Compute:

$\int_2^{\infty} \frac{1}{x \ln x}\, dx = \lim_{b \to \infty} \int_2^{b} \frac{1}{x \ln x}\, dx = \lim_{b \to \infty} \left[\ln(\ln x)\right]_2^b = \lim_{b \to \infty} \ln(\ln b) - \ln(\ln 2) = \infty$

The integral diverges, so by the integral test, the series diverges. $\blacksquare$

Worked Example: Approximate $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ to within $0.01$

Solution. This is the alternating harmonic series, with $a_n = 1/n$ . By the alternating series Estimation theorem, $|S - S_N| \leq a_{N+1} = 1/(N+1)$ . We need $1/(N+1) \leq 0.01$ So $N + 1 \geq 100$ I.e., $N \geq 99$ .

So $S_{99} = \sum_{n=1}^{99} \frac{(-1)^{n+1}}{n}$ approximates $\ln 2$ to within $0.01$ . (The exact sum is $\ln 2 \approx 0.6931$ .) $\blacksquare$

Worked Example: Determine convergence of $\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$

Solution. Since $\frac{1}{n^2 + 1} \leq \frac{1}{n^2}$ for all $n$ And $\sum 1/n^2$ converges ( $p$ -series with $p = 2 > 1$ ), the comparison test implies $\sum \frac{1}{n^2 + 1}$ converges. $\blacksquare$

Worked Example: Use the condensation test for $\sum_{n=2}^{\infty} \frac{1}{n (\ln n) (\ln \ln n)}$

Solution. Let $a_n = \frac{1}{n (\ln n)(\ln \ln n)}$ for $n \geq 3$ . This is positive and decreasing. By the condensation test, $\sum a_n$ converges iff $\sum 2^k a_{2^k}$ converges. Compute:

$2^k a_{2^k} = \frac{2^k}{2^k \cdot k \ln 2 \cdot \ln(k \ln 2)} = \frac{1}{k \ln 2 \cdot \ln(k \ln 2)} \approx \frac{1}{k \ln k}$

The series $\sum \frac{1}{k \ln k}$ diverges (integral test, analogous to $\sum \frac{1}{n \ln n}$ ). Therefore $\sum a_n$ diverges. $\blacksquare$

If you get this wrong, revise: Section 3.5 (Cauchy Condensation Test).

:::caution Common Pitfall The ratio and root tests are inconclusive when the limit equals 1. In such cases, try the comparison Test, integral test, or other methods. For example, $\sum 1/n$ diverges (harmonic series) and $\sum 1/n^2$ converges, but both give a ratio test limit of 1. :::

4. Continuity

4.1 Limits of Functions

Let $f : D \to \mathbb{R}$ where $D \subseteq \mathbb{R}$ . We say $\lim_{x \to a} f(x) = L$ if for Every $\varepsilon > 0$ There exists $\delta > 0$ such that

$0 \lt |x - a| \lt \delta \implies |f(x) - L| \lt \varepsilon$

4.2 Continuity

Definition. $f$ is continuous at $a$ if $\lim_{x \to a} f(x) = f(a)$ . In epsilon-delta form: For every $\varepsilon > 0$ There exists $\delta > 0$ such that

$|x - a| \lt \delta \implies |f(x) - f(a)| \lt \varepsilon$

Remark. A function is continuous on a set $E$ if it is continuous at every point of $E$ . A function is globally continuous (or “continuous”) if it is continuous on its entire domain.

Definition. $f$ is discontinuous at $a$ if it is not continuous at $a$ . Discontinuities are Classified as:

Removable: $\lim_{x \to a} f(x)$ exists but does not equal $f(a)$ (or $f(a)$ is undefined).
Jump: $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$ both exist but are unequal.
Essential (or infinite/oscillatory): At least one one-sided limit does not exist.

Proposition 4.3. Polynomials are continuous on $\mathbb{R}$ . Rational functions $p(x)/q(x)$ are Continuous wherever $q(x) \neq 0$ . The functions $\sin x$ , $\cos x$ , $e^x$ , $\ln x$ are continuous On their domains.

Theorem 4.1 (Algebra of Continuous Functions). If $f$ and $g$ are continuous at $a$ Then $f+g$ $f-g$ , $fg$ And (where defined) $f/g$ are continuous at $a$ .

Theorem 4.2. Compositions of continuous functions are continuous: if $f$ is continuous at $a$ and $g$ is continuous at $f(a)$ Then $g \circ f$ is continuous at $a$ .

4.2a Sequential Characterization of Limits and Continuity

The epsilon-delta definitions can be reformulated in terms of sequences, which is often more Convenient for proofs.

Proposition 4.2a (Sequential Criterion for Limits). $\lim_{x \to c} f(x) = L$ if and only if For every sequence $(x_n)$ with $x_n \to c$ and $x_n \neq c$ for all $n$ We have $f(x_n) \to L$ .

Proof. ( $\Rightarrow$ ) Let $\varepsilon > 0$ . Choose $\delta > 0$ from the $\varepsilon$ - $\delta$ definition. Since $x_n \to c$ There exists $N$ with $|x_n - c| \lt \delta$ for $n \geq N$ . Then $|f(x_n) - L| \lt \varepsilon$ for $n \geq N$ .

( $\Leftarrow$ ) Suppose the $\varepsilon$ - $\delta$ condition fails. Then there exists $\varepsilon > 0$ such That for every $n \in \mathbb{N}$ There exists $x_n$ with $0 \lt |x_n - c| \lt 1/n$ but $|f(x_n) - L| \geq \varepsilon$ . Then $x_n \to c$ but $f(x_n) \not\to L$ Contradicting the hypothesis. $\blacksquare$

Corollary 4.2b. $f$ is continuous at $c$ if and only if for every sequence $(x_n)$ with $x_n \to c$ We have $f(x_n) \to f(c)$ .

This is especially useful for proving that a function is not continuous: find one sequence Converging to $c$ whose image does not converge to $f(c)$ .

4.3 Intermediate Value Theorem

Theorem 4.3 (IVT). If $f : [a,b] \to \mathbb{R}$ is continuous and $f(a) \lt y \lt f(b)$ (or $f(b) \lt y \lt f(a)$ ), then there exists $c \in (a,b)$ such that $f(c) = y$ .

Proof. Assume $f(a) \lt y \lt f(b)$ . Let $S = \{x \in [a,b] : f(x) \lt y\}$ . Since $a \in S$ $S$ is non-empty and bounded above by $b$ . Let $c = \sup(S)$ . We show $f(c) = y$ .

If $f(c) \lt y$ Then by continuity at $c$ There exists $\delta > 0$ such that $f(x) \lt y$ for $x \in (c - \delta, c + \delta)$ . But then $c + \delta/2 \in S$ Contradicting that $c = \sup(S)$ .

If $f(c) > y$ Then by continuity, there exists $\delta > 0$ such that $f(x) > y$ for $x \in (c - \delta, c + \delta)$ . But then $c - \delta/2$ is an upper bound for $S$ Contradicting That $c = \sup(S)$ .

Therefore $f(c) = y$ . $\blacksquare$

Alternative proof (bisection). Set $a_0 = a$ , $b_0 = b$ . Given $[a_n, b_n]$ with $f(a_n) \lt y \lt f(b_n)$ Let $m_n = (a_n + b_n)/2$ . If $f(m_n) \geq y$ Set $a_{n+1} = a_n$ , $b_{n+1} = m_n$ . If $f(m_n) \lt y$ Set $a_{n+1} = m_n$ , $b_{n+1} = b_n$ . Either way, $f(a_n) \lt y \leq f(b_n)$ and $b_n - a_n = (b-a)/2^n \to 0$ . By the nested interval property, $a_n \to c$ and $b_n \to c$ . By continuity, $f(c) = \lim f(a_n) \leq y$ And $f(c) = \lim f(b_n) \geq y$ So $f(c) = y$ . $\blacksquare$

4.4 Extreme Value Theorem

Theorem 4.4 (EVT). If $f : [a,b] \to \mathbb{R}$ is continuous, then $f$ attains its maximum and Minimum on $[a,b]$ : there exist $c_1, c_2 \in [a,b]$ such that $f(c_1) \leq f(x) \leq f(c_2)$ for all $x \in [a,b]$ .

Proof. We first show $f$ is bounded. Suppose not; then for each $n \in \mathbb{N}$ There exists $x_n \in [a,b]$ with $|f(x_n)| > n$ . By Bolzano-Weierstrass, $(x_n)$ has a convergent subsequence $x_{n_k} \to c \in [a,b]$ . By continuity, $f(x_{n_k}) \to f(c)$ So $(f(x_{n_k}))$ is bounded. But $|f(x_{n_k})| > n_k \to \infty$ A contradiction.

Now we show $f$ attains its supremum. Let $M = \sup\{f(x) : x \in [a,b]\}$ . For each $n$ Choose $x_n \in [a,b]$ with $f(x_n) > M - 1/n$ . By Bolzano-Weierstrass, $(x_n)$ has a subsequence $x_{n_k} \to c \in [a,b]$ . By continuity, $f(c) = \lim f(x_{n_k})$ . Since $M - 1/n_k \lt f(x_{n_k}) \leq M$ for all $k$ The squeeze theorem gives $f(c) = M$ . The argument for the infimum is similar (consider $-f$ ). $\blacksquare$

4.5 Uniform Continuity

Definition. $f$ is uniformly continuous on $D$ if for every $\varepsilon > 0$ There exists $\delta > 0$ such that for all $x, y \in D$ :

$|x - y| \lt \delta \implies |f(x) - f(y)| \lt \varepsilon$

The key distinction: for ordinary continuity, $\delta$ may depend on both $\varepsilon$ and the point $a$ ; for uniform continuity, $\delta$ depends only on $\varepsilon$ .

4.6 The Heine-Cantor Theorem

Theorem 4.5 (Heine-Cantor). If $f : [a,b] \to \mathbb{R}$ is continuous on the closed, bounded Interval $[a,b]$ Then $f$ is uniformly continuous on $[a,b]$ .

Proof. Suppose $f$ is continuous on $[a,b]$ but not uniformly continuous. Then there exists $\varepsilon > 0$ such that for every $n \in \mathbb{N}$ There exist $x_n, y_n \in [a,b]$ with $|x_n - y_n| \lt 1/n$ but $|f(x_n) - f(y_n)| \geq \varepsilon$ .

By the Bolzano-Weierstrass theorem, $(x_n)$ has a convergent subsequence $x_{n_k} \to c \in [a,b]$ . Since $|x_{n_k} - y_{n_k}| \lt 1/n_k \to 0$ We have $y_{n_k} \to c$ as well.

By continuity of $f$ at $c$ : there exists $\delta > 0$ such that $|x - c| \lt \delta$ implies $|f(x) - f(c)| \lt \varepsilon/2$ . For $k$ sufficiently large, $|x_{n_k} - c| \lt \delta$ and $|y_{n_k} - c| \lt \delta$ So:

$|f(x_{n_k}) - f(y_{n_k})| \leq |f(x_{n_k}) - f(c)| + |f(y_{n_k}) - f(c)| \lt \varepsilon/2 + \varepsilon/2 = \varepsilon$

Contradicting $|f(x_{n_k}) - f(y_{n_k})| \geq \varepsilon$ . $\blacksquare$

4.7 Worked Examples

Problem. Prove that $f(x) = \sqrt{x}$ is uniformly continuous on $[0, \infty)$ .

Solution. For $x, y \geq 0$ : $|\sqrt{x} - \sqrt{y}| = \frac{|x - y|}{\sqrt{x} + \sqrt{y}} \leq |x - y|^{1/2}$ .

Given $\varepsilon > 0$ Choose $\delta = \varepsilon^2$ . Then $|x - y| \lt \delta$ implies $|\sqrt{x} - \sqrt{y}| \leq \sqrt{|x-y|} \lt \sqrt{\delta} = \varepsilon$ . Since $\delta$ depends Only on $\varepsilon$ The continuity is uniform. $\blacksquare$

Worked Example: $\varepsilon$-$\delta$ proof that $f(x) = 3x - 1$ is continuous at $x = 2$

Solution. We have $f(2) = 5$ . Let $\varepsilon > 0$ . We need to find $\delta > 0$ such that $|x - 2| \lt \delta$ implies $|f(x) - 5| \lt \varepsilon$ .

Compute: $|f(x) - 5| = |(3x - 1) - 5| = |3x - 6| = 3|x - 2|$ .

Choose $\delta = \varepsilon/3$ . Then $|x - 2| \lt \delta$ implies $|f(x) - 5| = 3|x - 2| \lt 3 \cdot \varepsilon/3 = \varepsilon$ . $\blacksquare$

Worked Example: $\varepsilon$-$\delta$ proof that $f(x) = x^2$ is continuous at $x = 3$

Solution. We have $f(3) = 9$ . Let $\varepsilon > 0$ . Compute:

$|f(x) - 9| = |x^2 - 9| = |x + 3| \cdot |x - 3|$

Restrict to $\delta \leq 1$ So $|x - 3| \lt 1$ means $2 \lt x \lt 4$ Giving $|x + 3| \lt 7$ .

Choose $\delta = \min(1, \varepsilon/7)$ . Then $|x - 3| \lt \delta$ implies:

$|x^2 - 9| = |x + 3| \cdot |x - 3| \lt 7 \cdot \frac{\varepsilon}{7} = \varepsilon$

$\blacksquare$

Worked Example: Show $f(x) = 1/x$ is NOT uniformly continuous on $(0, 1)$

Solution. We show the negation of uniform continuity. Take $\varepsilon = 1$ . For any $\delta > 0$ Choose $n \in \mathbb{N}$ with $1/n \lt \delta$ . Set $x = 1/n$ and $y = 1/(2n)$ . Then $|x - y| = 1/(2n) \lt 1/n \lt \delta$ But:

$|f(x) - f(y)| = \left|\frac{1}{1/n} - \frac{1}{1/(2n)}\right| = |n - 2n| = n \geq 1 = \varepsilon$

So no single $\delta$ works for all $x, y \in (0,1)$ . $\blacksquare$

Worked Example: Use the sequential criterion to show $f(x) = \sin(1/x)$ has no limit as $x \to 0$

Solution. Consider the sequences $x_n = 1/(2n\pi)$ and $y_n = 1/(2n\pi + \pi/2)$ . Both converge to $0$ . But $f(x_n) = \sin(2n\pi) = 0$ and $f(y_n) = \sin(2n\pi + \pi/2) = 1$ for all $n$ .

So $f(x_n) \to 0$ and $f(y_n) \to 1$ . By the sequential criterion, if $\lim_{x \to 0} f(x)$ existed, Both subsequences would converge to the same limit. Since they don”t, the limit does not exist. $\blacksquare$

Worked Example: Prove $f(x) = x \sin(1/x)$ (with $f(0) = 0$) is continuous everywhere

Solution. For $x \neq 0$ , $f$ is a product of continuous functions, hence continuous.

At $x = 0$ : let $\varepsilon > 0$ . Choose $\delta = \varepsilon$ . For $|x - 0| = |x| \lt \delta$ :

$|f(x) - f(0)| = |x \sin(1/x)| \leq |x| \lt \delta = \varepsilon$

So $f$ is continuous at $0$ . Since $f$ extends continuously from $(0, 1]$ to $[0, 1]$ The Heine-Cantor Theorem implies $f$ is uniformly continuous on $[0, 1]$ . $\blacksquare$

Worked Example: $\varepsilon$-$\delta$ proof that $f(x) = \sin x$ is continuous at every $a \in \mathbb{R}$

Solution. We use the identity $|\sin u - \sin v| \leq |u - v|$ for all $u, v \in \mathbb{R}$ . (Proof: $|\sin u - \sin v| = 2|\cos((u+v)/2)\sin((u-v)/2)| \leq 2|\sin((u-v)/2)| \leq |u - v|$ Using $|\sin t| \leq |t|$ and $|\cos| \leq 1$ .)

Let $\varepsilon > 0$ and $a \in \mathbb{R}$ . Choose $\delta = \varepsilon$ . For $|x - a| \lt \delta$ :

$|\sin x - \sin a| \leq |x - a| \lt \delta = \varepsilon$

Since $\delta = \varepsilon$ works independently of $a$ , $\sin x$ is actually uniformly continuous On $\mathbb{R}$ . The same argument works for $\cos x$ . $\blacksquare$

Worked Example: $\varepsilon$-$\delta$ proof that $f(x) = e^x$ is continuous at every $a \in \mathbb{R}$

Solution. We use the inequality $|e^u - e^v| \leq e^{\max(u,v)} |u - v|$ Which follows from the Mean Value Theorem applied to $e^t$ : $e^u - e^v = e^\xi (u - v)$ for some $\xi$ between $u$ and $v$ So $|e^u - e^v| = e^\xi |u - v| \leq e^{\max(u,v)} |u - v|$ .

Let $\varepsilon > 0$ and $a \in \mathbb{R}$ . Restrict to $|x - a| \lt 1$ So $x \lt a + 1$ and $e^{\max(x,a)} \leq e^{a+1}$ . Choose $\delta = \min(1, \varepsilon / e^{a+1})$ . For $|x - a| \lt \delta$ :

$|e^x - e^a| \leq e^{a+1} |x - a| \lt e^{a+1} \cdot \frac{\varepsilon}{e^{a+1}} = \varepsilon$

$\blacksquare$

If you get this wrong, revise: Section 4.2 (Continuity), Section 5.3 (Mean Value Theorem).

:::caution Common Pitfall Continuity on $(a, b)$ does not imply uniform continuity. The function $f(x) = 1/x$ on $(0, 1)$ is Continuous but not uniformly continuous. The Heine-Cantor theorem requires a closed and bounded Interval. Also, a function can be uniformly continuous on an unbounded domain (e.g., $f(x) = \sqrt{x}$ On $[0, \infty)$ ) --- boundedness of the domain is sufficient but not necessary. :::

5. Differentiability

5.1 The Derivative

Definition. $f : (a,b) \to \mathbb{R}$ is differentiable at $c \in (a,b)$ if the limit

$f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$

Exists (as a finite real number).

Proposition 5.1. If $f$ is differentiable at $c$ Then $f$ is continuous at $c$ .

Proof. $\lim_{x \to c} (f(x) - f(c)) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \cdot (x - c) = f'(c) \cdot 0 = 0$ . $\blacksquare$

The converse is false: $f(x) = |x|$ is continuous at $0$ but not differentiable at $0$ .

5.2 Differentiation Rules

Theorem 5.1. If $f$ and $g$ are differentiable at $c$ Then:

$(f + g)'(c) = f'(c) + g'(c)$
$(fg)'(c) = f'(c)g(c) + f(c)g'(c)$
$(f/g)'(c) = \frac{f'(c)g(c) - f(c)g'(c)}{g(c)^2}$ (if $g(c) \neq 0$ )
$(f \circ g)'(c) = f'(g(c)) \cdot g'(c)$ (Chain Rule)

5.3 Mean Value Theorem

Theorem 5.2 (Rolle’s Theorem). If $f : [a,b] \to \mathbb{R}$ is continuous on $[a,b]$ Differentiable On $(a,b)$ And $f(a) = f(b)$ Then there exists $c \in (a,b)$ such that $f'(c) = 0$ .

Proof. By the Extreme Value Theorem, $f$ attains its maximum $M$ and minimum $m$ on $[a,b]$ . If $M = m$ Then $f$ is constant and $f'(c) = 0$ for all $c \in (a,b)$ . Otherwise, at least one Of $M$ or $m$ is attained at some $c \in (a,b)$ (since $f(a) = f(b)$ ). By Fermat’s theorem, $f'(c) = 0$ . $\blacksquare$

Theorem 5.3 (Mean Value Theorem). If $f : [a,b] \to \mathbb{R}$ is continuous on $[a,b]$ and Differentiable on $(a,b)$ Then there exists $c \in (a,b)$ such that

$f'(c) = \frac{f(b) - f(a)}{b - a}$

Proof. Define $g(x) = f(x) - \frac{f(b)-f(a)}{b-a}(x - a)$ . Then $g(a) = g(b)$ and $g$ satisfies the Hypotheses of Rolle’s theorem. So $g'(c) = 0$ for some $c \in (a,b)$ Which gives the result. $\blacksquare$

Corollary 5.4. If $f'(x) = 0$ for all $x \in (a,b)$ Then $f$ is constant on $[a,b]$ .

Corollary 5.5. If $f'(x) > 0$ for all $x \in (a,b)$ Then $f$ is strictly increasing on $[a,b]$ .

Theorem 5.3a (Cauchy’s Mean Value Theorem). If $f, g : [a,b] \to \mathbb{R}$ are continuous on $[a,b]$ and differentiable on $(a,b)$ Then there exists $c \in (a,b)$ such that

$(f(b) - f(a))g'(c) = (g(b) - g(a))f'(c)$

Proof. Define $h(x) = (f(b) - f(a))g(x) - (g(b) - g(a))f(x)$ . Then $h(a) = h(b)$ So by Rolle’s Theorem, $h'(c) = 0$ for some $c \in (a,b)$ Which gives the result. $\blacksquare$

Remark. When $g(x) = x$ Cauchy’s MVT reduces to the standard MVT. Cauchy’s MVT is the key Ingredient in the proof of L’Hôpital’s rule.

Proof. Apply the MVT to $f$ on the interval between $x$ and $y$ . $\blacksquare$

5.4 Taylor’s Theorem

Theorem 5.6 (Taylor’s Theorem with Lagrange Remainder). If $f$ is $(n+1)$ -times differentiable on An open interval containing $a$ Then for each $x$ in that interval:

$f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x - a)^k + R_n(x)$

Where the remainder is

$R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x - a)^{n+1}$

For some $\xi$ between $a$ and $x$ .

Proof. Fix $x \neq a$ and define

$g(t) = f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(t)}{k!}(x - t)^k$

Then $g(a) = R_n(x)$ and $g(x) = 0$ . By the generalized Rolle’s theorem (or direct computation Using the Cauchy mean value theorem), there exists $\xi$ between $a$ and $x$ with $g'( \xi ) = 0$ . Computing:

$g'(t) = -\frac{f^{(n+1)}(t)}{n!}(x - t)^n$

Setting $g'(\xi) = 0$ yields the result after comparing $g(a) = R_n(x)$ with the integral form. A Cleaner approach uses the standard MVT applied to $g$ on $[a, x]$ . $\blacksquare$

5.5 L’Hôpital’s Rule

Theorem 5.7 (L’Hôpital’s Rule, $\frac{0}{0}$ case). Suppose $f$ and $g$ are differentiable on An open interval containing $c$ (except possibly at $c$ itself), $g'(x) \neq 0$ near $c$ And $\lim_{x \to c} f(x) = \lim_{x \to c} g(x) = 0$ . If $\lim_{x \to c} f'(x)/g'(x) = L$ exists (as a finite Number or $\pm\infty$ ), then $\lim_{x \to c} f(x)/g(x) = L$ .

Proof. Extend $f$ and $g$ continuously to $c$ by setting $f(c) = g(c) = 0$ . For $x \neq c$ By Cauchy’s Mean Value Theorem, there exists $\xi$ strictly between $c$ and $x$ such that

$\frac{f(x) - f(c)}{g(x) - g(c)} = \frac{f'(\xi)}{g'(\xi)}$

I.e., $\frac{f(x)}{g(x)} = \frac{f'(\xi)}{g'(\xi)}$ . As $x \to c$ We have $\xi \to c$ (since $\xi$ is trapped between $c$ and $x$ ). Therefore $\lim_{x \to c} f(x)/g(x) = \lim_{\xi \to c} f'(\xi)/g'(\xi) = L$ . $\blacksquare$

Theorem 5.7b (L’Hôpital’s Rule, $\frac{\infty}{\infty}$ case). Suppose $f$ and $g$ are Differentiable on $(a, b)$ (except possibly at $c$ ), $g'(x) \neq 0$ near $c$ And $\lim_{x \to c} |f(x)| = \lim_{x \to c} |g(x)| = \infty$ . If $\lim_{x \to c} f'(x)/g'(x) = L$ exists, Then $\lim_{x \to c} f(x)/g(x) = L$ .

Proof (sketch). Fix $\varepsilon > 0$ . For $x, y$ near $c$ with $x \neq y$ By Cauchy’s MVT:

$\frac{f(x) - f(y)}{g(x) - g(y)} = \frac{f'(\xi)}{g'(\xi)}$

For some $\xi$ between $x$ and $y$ . Since $f'(\xi)/g'(\xi) \approx L$ for $\xi$ near $c$ We have:

$\frac{f(x)}{g(x)} = \frac{f(x) - f(y)}{g(x) - g(y)} \cdot \frac{1 - f(y)/f(x)}{1 - g(y)/g(x)}$

Since $f(x), g(x) \to \infty$ By fixing $y$ and letting $x \to c$ The fractions $f(y)/f(x)$ and $g(y)/g(x)$ tend to $0$ So the second factor tends to $1$ . The first factor tends to $L$ by Cauchy’s MVT. Hence $f(x)/g(x) \to L$ . $\blacksquare$

Worked Example: Compute $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$

Solution. Both numerator and denominator approach $0$ as $x \to 0$ . Applying L’Hôpital’s rule:

$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x}$

This is still $\frac{0}{0}$ So apply L’Hôpital again:

$= \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}$

$\blacksquare$

5.6 Darboux’s Theorem

Theorem 5.8 (Darboux’s Theorem). If $f$ is differentiable on $[a, b]$ Then $f'$ has the Intermediate value property: for any $y$ between $f'(a)$ and $f'(b)$ There exists $c \in (a, b)$ With $f'(c) = y$ .

Remark. This means derivatives satisfy the intermediate value property even though they need not Be continuous. For example, $f(x) = x^2 \sin(1/x)$ (with $f(0) = 0$ ) is differentiable everywhere, But $f'$ is not continuous at $0$ .

Proof. Assume without loss of generality that $f'(a) \lt y \lt f'(b)$ . Define $g(x) = f(x) - yx$ . Then $g$ is differentiable on $[a, b]$ with

$g'(a) = f'(a) - y \lt 0 \quad \mathrm{and} \quad g'(b) = f'(b) - y > 0$

Since $g'(a) \lt 0$ There exists $x_1 > a$ with $g(x_1) \lt g(a)$ (otherwise $g(x) \geq g(a)$ For $x$ near $a$ Contradicting $g'(a) \lt 0$ ). Similarly, since $g'(b) > 0$ There exists $x_2 \lt b$ with $g(x_2) \lt g(b)$ .

Therefore $g$ attains its minimum at some $c \in (a, b)$ . By Fermat’s theorem on interior extrema, $g'(c) = 0$ So $f'(c) = y$ . $\blacksquare$

Worked Example: Apply Darboux's theorem to $f(x) = x^2 \sin(1/x)$ ($f(0) = 0$)

Solution. For $x \neq 0$ : $f'(x) = 2x \sin(1/x) - \cos(1/x)$ . At $x = 0$ : $f'(0) = \lim_{h \to 0} \frac{h^2 \sin(1/h)}{h} = \lim_{h \to 0} h \sin(1/h) = 0$ .

So $f'(0) = 0$ . For any $\delta > 0$ The term $-\cos(1/x)$ oscillates between $-1$ and $1$ on $(0, \delta)$ So $f'$ takes all values in $[-1, 1]$ infinitely often on $(0, \delta)$ .

But Darboux’s theorem says $f'$ has the intermediate value property. Indeed, $f'$ is not continuous At $0$ (it oscillates wildly), yet it still satisfies the IVP. This shows that derivatives can be Highly discontinuous while retaining the intermediate value property. $\blacksquare$

5.7 Worked Examples

Worked Example. Compute the third-order Taylor polynomial of $f(x) = e^x$ about $a = 0$ .

$f(0) = 1$ , $f'(0) = 1$ , $f''(0) = 1$ , $f'''(0) = 1$ . So

$T_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$

The remainder is $R_3(x) = \frac{e^\xi}{24} x^4$ for some $\xi$ between $0$ and $x$ .

Worked Example: Approximate $\sin(0.1)$ with error less than $10^{-10}$

Solution. For $f(x) = \sin x$ about $a = 0$ : $f^{(k)}(0) \in \{0, 1, -1\}$ and the Taylor Polynomial of degree $n$ has the form $T_n(x) = x - x^3/3! + x^5/5! - \cdots$ (odd terms only).

The Lagrange remainder is $|R_n(x)| = \frac{|f^{(n+1)}(\xi)|}{(n+1)!} |x|^{n+1} \leq \frac{|x|^{n+1}}{(n+1)!}$ (since $|f^{(k)}| \leq 1$ for all $k$ ).

We need $\frac{(0.1)^{n+1}}{(n+1)!} \lt 10^{-10}$ . Testing: for $n = 5$ $\frac{(0.1)^6}{6!} = \frac{10^{-6}}{720} \approx 1.39 \times 10^{-9} \gt 10^{-10}$ . For $n = 7$ : $\frac{(0.1)^8}{8!} = \frac{10^{-8}}{40320} \approx 2.48 \times 10^{-13} \lt 10^{-10}$ .

So $T_7(0.1) = 0.1 - \frac{(0.1)^3}{6} + \frac{(0.1)^5}{120} - \frac{(0.1)^7}{5040}$ $= 0.1 - 0.00016667 + 0.00000083 - 0.00000000 \approx 0.09983342$ .

The error is at most $2.48 \times 10^{-13} \lt 10^{-10}$ . $\blacksquare$

Worked Example: Find the Maclaurin series for $\ln(1 + x)$ and its radius of convergence

Solution. For $f(x) = \ln(1+x)$ : $f(0) = 0$ , $f'(x) = 1/(1+x)$ , $f''(x) = -1/(1+x)^2$ $f^{(k)}(x) = (-1)^{k-1}(k-1)!/(1+x)^k$ for $k \geq 1$ . So $f^{(k)}(0) = (-1)^{k-1}(k-1)!$ .

$\ln(1+x) = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}(k-1)!}{k!} x^k = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} x^k$

By the ratio test: $\lim_{k \to \infty} |a_{k+1}/a_k| = \lim_{k \to \infty} \frac{k}{k+1} |x| = |x|$ . The series converges for $|x| \lt 1$ and diverges for $|x| > 1$ . At $x = 1$ we get the alternating Harmonic series (converges to $\ln 2$ ). At $x = -1$ we get the negative harmonic series (diverges).

The radius of convergence is $R = 1$ and the interval of convergence is $(-1, 1]$ . $\blacksquare$

Worked Example: Compute the Taylor expansion of $\cos x$ about $a = \pi/3$ with remainder bound

Solution. Compute derivatives: $f(x) = \cos x$ , $f'(x) = -\sin x$ , $f''(x) = -\cos x$ , $f'''(x) = \sin x$ $f^{(4)}(x) = \cos x$ . Evaluated at $a = \pi/3$ :

$f(\pi/3) = 1/2$ , $f'(\pi/3) = -\sqrt{3}/2$ , $f''(\pi/3) = -1/2$ , $f'''(\pi/3) = \sqrt{3}/2$ .

The third-degree Taylor polynomial is:

$T_3(x) = \frac{1}{2} - \frac{\sqrt{3}}{2}\left(x - \frac{\pi}{3}\right) - \frac{1}{4}\left(x - \frac{\pi}{3}\right)^2 + \frac{\sqrt{3}}{12}\left(x - \frac{\pi}{3}\right)^3$

The remainder satisfies $|R_3(x)| \leq \frac{|x - \pi/3|^4}{24}$ (since $|f^{(4)}(\xi)| = |\cos \xi| \leq 1$ ).

For example, at $x = 1$ : $|R_3(1)| \leq \frac{|1 - \pi/3|^4}{24} \approx \frac{0.0472^4}{24} \approx 2.1 \times 10^{-7}$ . $\blacksquare$

:::caution Common Pitfall L’Hôpital’s rule only applies to indeterminate forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$ . Applying it to forms like $\frac{1}{0}$ or $\frac{\infty}{1}$ will give incorrect results. Always Verify the indeterminate form before applying the rule. Also, L’Hôpital’s rule requires that the Limit of the quotient of derivatives exists; if it does not exist (oscillates), the original limit May still exist. :::

6. Riemann Integration

6.1 Definition

Let $f : [a,b] \to \mathbb{R}$ be bounded. A partition of $[a,b]$ is a finite set $P = \{x_0, x_1, \ldots, x_n\}$ with $a = x_0 \lt x_1 \lt \cdots \lt x_n = b$ .

The upper sum and lower sum of $f$ with respect to $P$ are:

$U(f, P) = \sum_{i=1}^{n} M_i \Delta x_i, \quad L(f, P) = \sum_{i=1}^{n} m_i \Delta x_i$

Where $M_i = \sup\{f(x) : x \in [x_{i-1}, x_i]\}$ , $m_i = \inf\{f(x) : x \in [x_{i-1}, x_i]\}$ And $\Delta x_i = x_i - x_{i-1}$ .

The mesh of $P$ is $\|P\| = \max_{1 \leq i \leq n} \Delta x_i$ .

Definition. $f$ is Riemann integrable on $[a,b]$ if the upper and lower integrals are equal:

$\overline{\int_a^b} f(x)\, dx = \underline{\int_a^b} f(x)\, dx$

Where $\overline{\int_a^b} f = \inf\{U(f,P) : P \mathrm{\ is\ a\ partition}\}$ and $\underline{\int_a^b} f = \sup\{L(f,P) : P \mathrm{\ is\ a\ partition}\}$ .

The common value is denoted $\int_a^b f(x)\, dx$ .

6.2 Integrability Criteria

Theorem 6.1 (Riemann Integrability Criterion). A bounded function $f : [a,b] \to \mathbb{R}$ is Riemann integrable if and only if for every $\varepsilon > 0$ There exists a partition $P$ such that

$U(f,P) - L(f,P) \lt \varepsilon$

Theorem 6.2. Every continuous function on $[a,b]$ is Riemann integrable.

Proof. Let $f$ be continuous on $[a,b]$ . By the Heine-Cantor theorem, $f$ is uniformly continuous. Given $\varepsilon > 0$ Choose $\delta > 0$ such that $|x - y| \lt \delta$ implies $|f(x) - f(y)| \lt \varepsilon/(b-a)$ .

Let $P$ be any partition with $\|P\| \lt \delta$ . On each subinterval $[x_{i-1}, x_i]$ By the Extreme Value Theorem, $f$ attains its maximum $M_i$ and minimum $m_i$ . By uniform continuity: $M_i - m_i \lt \varepsilon/(b-a)$ . Therefore:

$U(f,P) - L(f,P) = \sum_{i=1}^{n}(M_i - m_i)\Delta x_i \lt \frac{\varepsilon}{b-a} \sum_{i=1}^{n} \Delta x_i = \varepsilon$

By the Riemann integrability criterion, $f$ is integrable. $\blacksquare$

Theorem 6.3. Every monotone function on $[a,b]$ is Riemann integrable.

Proof. Assume $f$ is increasing (the decreasing case is analogous). Given $\varepsilon > 0$ Let $P_n$ be the uniform partition with $n$ subintervals of length $(b-a)/n$ . On $[x_{i-1}, x_i]$ : $M_i = f(x_i)$ and $m_i = f(x_{i-1})$ . Then:

$U(f, P_n) - L(f, P_n) = \sum_{i=1}^{n} [f(x_i) - f(x_{i-1})] \cdot \frac{b-a}{n} = [f(b) - f(a)] \cdot \frac{b-a}{n}$

Choose $n$ large enough that $[f(b) - f(a)](b-a)/n \lt \varepsilon$ . $\blacksquare$

Theorem 6.4. A bounded function with finitely many discontinuities on $[a,b]$ is Riemann integrable.

Proof (sketch). Let $f$ have discontinuities at $d_1, \ldots, d_m \in [a,b]$ . Given $\varepsilon > 0$ Enclose each $d_j$ in a small interval $I_j$ of total length $\varepsilon/(2M)$ Where $M = \sup_{[a,b]} |f|$ . On the remaining set (a finite union of closed intervals), $f$ is continuous, Hence uniformly continuous. Choose a partition fine enough that the oscillation of $f$ on each Subinterval outside the $I_j$ is less than $\varepsilon/(2(b-a))$ . Then:

$U(f, P) - L(f, P) \leq \frac{\varepsilon}{2(b-a)} \cdot (b - a) + 2M \cdot \frac{\varepsilon}{2M} = \varepsilon$

$\blacksquare$

Proposition 6.4a. The set of Riemann integrable functions on $[a,b]$ forms a vector space, and If $f$ and $g$ are integrable, then so are $|f|$ , $f^2$ And $\max(f, g)$ .

Theorem 6.4b (Lebesgue’s Criterion for Riemann Integrability). A bounded function $f : [a,b] \to \mathbb{R}$ Is Riemann integrable if and only if the set of its discontinuities has (Lebesgue) measure zero.

Remark. A set has measure zero if it can be covered by countably many intervals of arbitrarily Small total length. In particular, every countable set has measure zero. This means:

Every continuous function is integrable (empty set of discontinuities).
Every function with countably many discontinuities is integrable (Theorem 6.4 is a special case).
The Dirichlet function $f(x) = 1$ for $x \in \mathbb{Q}$ and $f(x) = 0$ for $x \notin \mathbb{Q}$ is discontinuous everywhere (set of discontinuities = $[a,b]$ Measure $> 0$ ), hence not integrable.
Thomae’s function $f(x) = 1/q$ if $x = p/q$ in lowest terms, and $f(x) = 0$ if $x$ is irrational, is continuous at every irrational and discontinuous at every rational. Since $\mathbb{Q}$ is countable (measure zero), Thomae’s function is Riemann integrable, with $\int_0^1 f = 0$ .

6.3 Properties of the Integral

Theorem 6.5 (Linearity). If $f$ and $g$ are integrable on $[a,b]$ and $\alpha, \beta \in \mathbb{R}$ :

$\int_a^b (\alpha f + \beta g) = \alpha \int_a^b f + \beta \int_a^b g$

Theorem 6.6 (Monotonicity). If $f(x) \leq g(x)$ for all $x \in [a,b]$ Then $\int_a^b f \leq \int_a^b g$ .

Theorem 6.7 (Triangle Inequality). $\left|\int_a^b f\right| \leq \int_a^b |f|$ .

6.4 The Fundamental Theorem of Calculus

Theorem 6.8 (FTC Part 1). If $f$ is continuous on $[a,b]$ Then the function

$F(x) = \int_a^x f(t)\, dt$

Is differentiable on $(a,b)$ and $F'(x) = f(x)$ .

Proof. Let $h > 0$ (the case $h \lt 0$ is similar). By the Mean Value Theorem for Integrals (which follows from the EVT), there exists $\xi \in [x, x+h]$ such that

$\frac{F(x+h) - F(x)}{h} = \frac{1}{h}\int_x^{x+h} f(t)\, dt = f(\xi)$

As $h \to 0^+$ We have $\xi \to x^+$ (since $\xi \in [x, x+h]$ ). By continuity of $f$ $f(\xi) \to f(x)$ . Hence $F'_+(x) = f(x)$ . A similar argument gives $F'_-(x) = f(x)$ . $\blacksquare$

Theorem 6.9 (FTC Part 2). If $F$ is differentiable on $[a,b]$ with $F' = f$ (and $f$ is integrable), Then

$\int_a^b f(x)\, dx = F(b) - F(a)$

Proof. Let $P = \{x_0, \ldots, x_n\}$ be any partition of $[a,b]$ . By the Mean Value Theorem, For each $i$ there exists $\xi_i \in [x_{i-1}, x_i]$ with $F(x_i) - F(x_{i-1}) = f(\xi_i)\Delta x_i$ . Summing:

$F(b) - F(a) = \sum_{i=1}^{n} [F(x_i) - F(x_{i-1})] = \sum_{i=1}^{n} f(\xi_i) \Delta x_i$

The right-hand side is a Riemann sum for $\int_a^b f$ . As $\|P\| \to 0$ This converges to the Integral. Hence $F(b) - F(a) = \int_a^b f(x)\, dx$ . $\blacksquare$

6.5 Worked Examples

Problem. Compute $\int_0^1 x^2\, dx$ from the definition.

Solution. Let $P_n = \{0, 1/n, 2/n, \ldots, 1\}$ . On $[x_{i-1}, x_i] = [(i-1)/n, i/n]$ , $f(x) = x^2$ Has $M_i = (i/n)^2$ and $m_i = ((i-1)/n)^2$ .

$U(f, P_n) = \sum_{i=1}^{n} \frac{i^2}{n^2} \cdot \frac{1}{n} = \frac{1}{n^3} \sum_{i=1}^{n} i^2 = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}$

As $n \to \infty$ : $\lim_{n \to \infty} U(f, P_n) = \lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} = \frac{2}{6} = \frac{1}{3}$ .

Similarly, $L(f, P_n) \to 1/3$ . So $\int_0^1 x^2\, dx = 1/3$ . $\blacksquare$

Worked Example: Compute $\int_0^1 \sqrt{x}\, dx$ from the definition

Solution. Let $P_n = \{0, 1/n, 2/n, \ldots, 1\}$ . On $[(i-1)/n, i/n]$ , $f(x) = \sqrt{x}$ has $M_i = \sqrt{i/n}$ and $m_i = \sqrt{(i-1)/n}$ .

$U(f, P_n) = \sum_{i=1}^{n} \sqrt{\frac{i}{n}} \cdot \frac{1}{n} = \frac{1}{n^{3/2}} \sum_{i=1}^{n} \sqrt{i}$

Using $\sum_{i=1}^{n} \sqrt{i} = \frac{2}{3} n^{3/2} + O(n^{1/2})$ (obtained from comparing with $\int_0^n \sqrt{x}\, dx$ ):

$\lim_{n \to \infty} U(f, P_n) = \lim_{n \to \infty} \frac{1}{n^{3/2}} \cdot \frac{2}{3}n^{3/2} = \frac{2}{3}$

Similarly $L(f, P_n) \to 2/3$ Confirming $\int_0^1 \sqrt{x}\, dx = 2/3$ . $\blacksquare$

6.6 Improper Integrals

Definition. An improper integral is a Riemann integral where either the interval of integration Is unbounded or the integrand is unbounded.

Type I (Infinite Intervals). If $f$ is Riemann integrable on $[a, b]$ for every $b > a$ Define:

$\int_a^{\infty} f(x)\, dx = \lim_{b \to \infty} \int_a^b f(x)\, dx$

The integral converges if this limit exists as a finite number; otherwise it diverges.

Type II (Unbounded Integrands). If $f$ is unbounded near $a$ but integrable on $[c, b]$ for every $c \in (a, b]$ :

$\int_a^b f(x)\, dx = \lim_{c \to a^+} \int_c^b f(x)\, dx$

Theorem 6.10 (Comparison Test for Improper Integrals). If $0 \leq f(x) \leq g(x)$ for $x \geq a$ :

If $\int_a^{\infty} g$ converges, then $\int_a^{\infty} f$ converges.
If $\int_a^{\infty} f$ diverges, then $\int_a^{\infty} g$ diverges.

Theorem 6.11 (Absolute Convergence). If $\int_a^{\infty} |f(x)|\, dx$ converges, then $\int_a^{\infty} f(x)\, dx$ converges.

Theorem 6.12 ( $p$ -Test for Improper Integrals).

Type I: $\int_1^{\infty} \frac{1}{x^p}\, dx$ converges if and only if $p > 1$ .
Type II: $\int_0^1 \frac{1}{x^p}\, dx$ converges if and only if $p < 1$ .

Proof. For Type I with $p \neq 1$ :

$\int_1^{\infty} x^{-p}\, dx = \lim_{b \to \infty} \left[\frac{x^{1-p}}{1-p}\right]_1^b = \lim_{b \to \infty} \frac{b^{1-p} - 1}{1-p}$

This converges when $1 - p < 0$ I.e., $p > 1$ . For $p = 1$ : $\int_1^{\infty} 1/x\, dx = \lim_{b \to \infty} \ln b = \infty$ .

For Type II: $\int_0^1 x^{-p}\, dx = \lim_{c \to 0^+} \frac{1 - c^{1-p}}{1-p}$ . This converges when $1 - p > 0$ I.e., $p < 1$ . $\blacksquare$

Remark. The $p$ -test for Type I integrals mirrors the $p$ -series test: $\sum 1/n^p$ converges Iff $p > 1$ . This is not a coincidence --- the integral test establishes the connection.

Worked Example: Evaluate $\int_0^{\infty} e^{-x}\, dx$

Solution. This is a Type I improper integral:

$\int_0^{\infty} e^{-x}\, dx = \lim_{b \to \infty} \int_0^b e^{-x}\, dx = \lim_{b \to \infty} \left[-e^{-x}\right]_0^b = \lim_{b \to \infty} (-e^{-b} + 1) = 1$

So the integral converges to $1$ . $\blacksquare$

Worked Example: Does $\int_1^{\infty} \frac{\sin x}{x}\, dx$ converge?

Solution. The integral $\int_1^{\infty} \left|\frac{\sin x}{x}\right|\, dx$ diverges (compare with $\int_1^{\infty} \frac{|\sin x|}{x}\, dx \geq \sum_{k=1}^{\infty} \int_{k\pi}^{(k+1)\pi} \frac{|\sin x|}{x}\, dx \geq \sum_{k=1}^{\infty} \frac{2}{(k+1)\pi}$ , which diverges by comparison with the harmonic series).

However, $\int_1^{\infty} \frac{\sin x}{x}\, dx$ converges by Dirichlet’s test for integrals. Let $F(b) = \int_1^b \sin x\, dx = \cos 1 - \cos b$ Which is bounded by $|\cos 1 - \cos b| \leq 2$ . Since $1/x$ decreases to $0$ By integration by parts:

$\int_1^b \frac{\sin x}{x}\, dx = \frac{-\cos x}{x}\bigg|_1^b - \int_1^b \frac{\cos x}{x^2}\, dx$

As $b \to \infty$ The boundary term $\cos b / b \to 0$ and $\int_1^{\infty} \frac{|\cos x|}{x^2}\, dx \leq \int_1^{\infty} \frac{1}{x^2}\, dx = 1$ , so the improper integral converges (conditionally). $\blacksquare$

Worked Example: Evaluate $\int_0^1 \frac{1}{\sqrt{x}}\, dx$ (Type II improper integral)

Solution. The integrand $f(x) = 1/\sqrt{x}$ is unbounded as $x \to 0^+$ . Compute:

$\int_0^1 \frac{1}{\sqrt{x}}\, dx = \lim_{c \to 0^+} \int_c^1 x^{-1/2}\, dx = \lim_{c \to 0^+} \left[2\sqrt{x}\right]_c^1 = \lim_{c \to 0^+} (2 - 2\sqrt{c}) = 2$

The improper integral converges to $2$ . Note that $\int_0^1 x^{-p}\, dx$ converges for $p \lt 1$ and Diverges for $p \geq 1$ . $\blacksquare$

Worked Example: Determine convergence of $\int_1^{\infty} \frac{1}{x^p}\, dx$ for various $p$

Solution. By the $p$ -test (Theorem 6.12): $\int_1^{\infty} x^{-p}\, dx$ converges iff $p > 1$ .

Specifically:

$p = 2$ : $\int_1^{\infty} 1/x^2\, dx = \lim_{b \to \infty} [-1/x]_1^b = 0 - (-1) = 1$ . Converges.
$p = 1$ : $\int_1^{\infty} 1/x\, dx = \lim_{b \to \infty} \ln b = \infty$ . Diverges.
$p = 1/2$ : $\int_1^{\infty} 1/\sqrt{x}\, dx = \lim_{b \to \infty} [2\sqrt{x}]_1^b = \infty$ . Diverges.

This mirrors the $p$ -series test: $\sum 1/n^p$ converges iff $p > 1$ . $\blacksquare$

Worked Example: Evaluate $\int_0^{\infty} x e^{-x}\, dx$

Solution. This integral requires both a Type I and Type II limit:

$\int_0^{\infty} x e^{-x}\, dx = \lim_{a \to 0^+} \lim_{b \to \infty} \int_a^b x e^{-x}\, dx$

Integrate by parts with $u = x$ , $dv = e^{-x}\, dx$ So $du = dx$ , $v = -e^{-x}$ :

$\int x e^{-x}\, dx = -xe^{-x} + \int e^{-x}\, dx = -xe^{-x} - e^{-x} = -(x+1)e^{-x}$

Evaluating: $\lim_{b \to \infty} [-(b+1)e^{-b}] - \lim_{a \to 0^+} [-(a+1)e^{-a}] = 0 - (-1) = 1$ .

So $\int_0^{\infty} x e^{-x}\, dx = 1$ . This equals $\Gamma(2) = 1! = 1$ . $\blacksquare$

:::caution Common Pitfall The Riemann integral is defined for bounded functions on closed, bounded intervals. For unbounded Functions or infinite intervals, one must use the improper Riemann integral. A common error is Applying the FTC directly to improper integrals without taking the limit. Also, conditional Convergence of improper integrals behaves differently from absolute convergence: rearranging the “terms” (subintervals) of a conditionally convergent improper integral can change its value. :::

7. Sequences and Series of Functions

7.1 Pointwise Convergence

Let $(f_n)$ be a sequence of functions defined on a set $E \subseteq \mathbb{R}$ .

Definition. $(f_n)$ converges pointwise to $f$ on $E$ if for every $x \in E$ and every $\varepsilon > 0$ There exists $N \in \mathbb{N}$ (depending on both $x$ and $\varepsilon$ ) such that $|f_n(x) - f(x)| \lt \varepsilon$ for all $n \geq N$ .

Example. Let $f_n(x) = x^n$ on $E = [0, 1]$ . For each $x \in [0, 1)$ , $f_n(x) = x^n \to 0$ And $f_n(1) = 1$ for all $n$ . So $f_n$ converges pointwise to

$f(x) = \begin{cases} 0 & \mathrm{if\ } 0 \leq x \lt 1 \\ 1 & \mathrm{if\ } x = 1 \end{cases}$

Note that each $f_n$ is continuous, but the pointwise limit $f$ is not continuous at $x = 1$ .

7.2 Uniform Convergence

Definition. $(f_n)$ converges uniformly to $f$ on $E$ if for every $\varepsilon > 0$ There Exists $N \in \mathbb{N}$ (depending only on $\varepsilon$ Not on $x$ ) such that for all $x \in E$ :

$|f_n(x) - f(x)| \lt \varepsilon \quad \mathrm{for\ all\ } n \geq N$

Equivalently, $\sup_{x \in E} |f_n(x) - f(x)| \to 0$ as $n \to \infty$ .

Proposition 7.1. Uniform convergence implies pointwise convergence. The converse is false.

Example (continued). $f_n(x) = x^n$ on $[0, 1]$ converges pointwise but not uniformly. We have $\sup_{x \in [0,1]} |f_n(x) - f(x)| = \sup_{x \in [0,1)} x^n = 1$ for all $n$ (since the supremum is Approached as $x \to 1^-$ ). This does not tend to $0$ .

However, on $[0, r]$ for any $r \lt 1$ : $\sup_{x \in [0,r]} |x^n| = r^n \to 0$ So the convergence Is uniform on $[0, r]$ .

7.3 The Weierstrass M-Test

Theorem 7.1 (Weierstrass M-Test). Let $(f_n)$ be a sequence of functions on $E$ . If there exists a Sequence $(M_n)$ of non-negative real numbers such that $|f_n(x)| \leq M_n$ for all $x \in E$ and all $n$ And $\sum_{n=1}^{\infty} M_n \lt \infty$ Then $\sum_{n=1}^{\infty} f_n$ converges uniformly on $E$ .

Proof. Let $S_n(x) = \sum_{k=1}^{n} f_k(x)$ and $T_n = \sum_{k=1}^{n} M_k$ . Since $\sum M_k$ converges, $(T_n)$ is a Cauchy sequence. Given $\varepsilon > 0$ There exists $N$ such that for $m > n \geq N$ :

$T_m - T_n = \sum_{k=n+1}^{m} M_k \lt \varepsilon$

Then for all $x \in E$ and $m > n \geq N$ :

$|S_m(x) - S_n(x)| = \left|\sum_{k=n+1}^{m} f_k(x)\right| \leq \sum_{k=n+1}^{m} |f_k(x)| \leq \sum_{k=n+1}^{m} M_k \lt \varepsilon$

So the partial sums $(S_n)$ satisfy the uniform Cauchy criterion on $E$ Hence converge uniformly. $\blacksquare$

7.4 Uniform Convergence and Continuity

Theorem 7.2. If $(f_n)$ is a sequence of continuous functions on $E$ converging uniformly to $f$ On $E$ Then $f$ is continuous on $E$ .

$|f(x) - f(c)| \leq |f(x) - f_N(x)| + |f_N(x) - f_N(c)| + |f_N(c) - f(c)| \lt \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon$

$\blacksquare$

7.5 Uniform Convergence and Integration

Theorem 7.3. If $(f_n)$ is a sequence of Riemann integrable functions on $[a, b]$ converging Uniformly to $f$ on $[a, b]$ Then $f$ is Riemann integrable and

$\lim_{n \to \infty} \int_a^b f_n(x)\, dx = \int_a^b f(x)\, dx$

Proof. Since $(f_n)$ converges uniformly, $f$ is the uniform limit of integrable functions. Given $\varepsilon > 0$ Choose $N$ with $\sup |f_N(x) - f(x)| \lt \varepsilon/(2(b-a))$ for all $x \in [a, b]$ . Then $f_N - \varepsilon/(2(b-a)) \leq f(x) \leq f_N(x) + \varepsilon/(2(b-a))$ for all $x$ And by Integrability of $f_N$ :

$\int_a^b f_N - \frac{\varepsilon}{2} \leq \underline{\int_a^b} f \leq \overline{\int_a^b} f \leq \int_a^b f_N + \frac{\varepsilon}{2}$

So $\overline{\int} f - \underline{\int} f \leq \varepsilon$ Proving $f$ is integrable. For the limit:

$\left|\int_a^b f_n - \int_a^b f\right| \leq \int_a^b |f_n - f| \leq (b-a) \cdot \sup_{[a,b]} |f_n - f| \to 0$

$\blacksquare$

7.6 Uniform Convergence and Differentiation

Uniform convergence of functions does not guarantee convergence of derivatives. A stronger Hypothesis is needed.

Theorem 7.4. Suppose $(f_n)$ is a sequence of differentiable functions on $[a, b]$ such that:

$(f_n(c))$ converges for some $c \in [a, b]$
$(f_n')$ converges uniformly on $[a, b]$

Then $(f_n)$ converges uniformly to a differentiable function $f$ on $[a, b]$ And $f'(x) = \lim_{n \to \infty} f_n'(x)$ .

Proof. Let $g = \lim f_n'$ (uniform limit). Define $f(x) = \lim_{n \to \infty} \left[f_n(c) + \int_c^x f_n'(t)\, dt\right]$ . By Theorem 7.3, $\int_c^x f_n'(t)\, dt \to \int_c^x g(t)\, dt$ So $f(x) = f(c) + \int_c^x g(t)\, dt$ . By FTC Part 1, $f$ is differentiable and $f'(x) = g(x)$ . Uniform convergence of $f_n$ to $f$ follows From the estimate $|f_n(x) - f(x)| \leq |f_n(c) - f(c)| + \int_a^b |f_n'(t) - g(t)|\, dt$ . $\blacksquare$

7.7 Power Series

A power series centered at $a$ is a series of the form $\sum_{n=0}^{\infty} c_n (x - a)^n$ .

Theorem 7.5 (Radius of Convergence). Every power series $\sum c_n (x - a)^n$ has a radius of Convergence $R \in [0, \infty]$ such that:

The series converges absolutely for $|x - a| \lt R$
The series diverges for $|x - a| > R$
The behavior at $|x - a| = R$ must be checked separately

The radius is given by $1/R = \limsup_{n \to \infty} \sqrt[n]{|c_n|}$ (Cauchy-Hadamard formula), or When the limit exists, $R = \lim_{n \to \infty} |c_n/c_{n+1}|$ .

Proof. Apply the root test to $\sum |c_n (x-a)^n|$ : $\limsup \sqrt[n]{|c_n|} |x-a| = |x-a|/R$ (where $1/R = \limsup \sqrt[n]{|c_n|}$ ). The root test gives convergence when $|x-a|/R \lt 1$ And divergence when $|x-a|/R > 1$ . $\blacksquare$

Theorem 7.6. A power series converges uniformly on every compact subset of its open disk of Convergence.

Theorem 7.6a (Differentiation and Integration of Power Series). If $f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n$ Has radius of convergence $R > 0$ Then:

$f$ is differentiable on $(a - R, a + R)$ and $f'(x) = \sum_{n=1}^{\infty} n c_n (x - a)^{n-1}$ (same $R$ ).
$f$ is infinitely differentiable on $(a - R, a + R)$ And $f^{(k)}(x) = \sum_{n=k}^{\infty} \frac{n!}{(n-k)!} c_n (x - a)^{n-k}$ .
$\int_a^x f(t)\, dt = \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x - a)^{n+1}$ for $|x - a| \lt R$ .
$c_n = f^{(n)}(a)/n!$ (uniqueness of power series coefficients).

Proof. The differentiated series $\sum n c_n (x-a)^{n-1}$ has the same radius of convergence as The original (by the Cauchy-Hadamard formula, since $\sqrt[n]{n} \to 1$ ). By Theorem 7.4, the Derivative of the sum equals the sum of the derivatives. Parts (2), (3), and (4) follow by Induction and the FTC. $\blacksquare$

Theorem 7.6b (Abel’s Theorem). If $\sum_{n=0}^{\infty} c_n$ converges to $L$ Then

$\lim_{x \to 1^-} \sum_{n=0}^{\infty} c_n x^n = L$

That is, the power series is continuous from the left at the endpoint $x = 1$ .

Proof (sketch). Let $s_n = \sum_{k=0}^{n} c_k$ and $s_n \to L$ . Write the partial sum $\sum_{k=0}^{n} c_k x^k = \sum_{k=0}^{n}(s_k - s_{k-1})x^k$ (with $s_{-1} = 0$ ) and use summation by Parts to express this as $s_n x^n + \sum_{k=0}^{n-1} s_k(x^k - x^{k+1})$ . Letting $n \to \infty$ and using That $s_n \to L$ and $x^n \to 0$ for $|x| \lt 1$ One shows the expression tends to $L$ as $x \to 1^-$ . $\blacksquare$

Example. Since $\sum_{k=1}^{\infty} (-1)^{k+1}/k = \ln 2$ Abel’s theorem gives $\lim_{x \to 1^-} \sum_{k=1}^{\infty} (-1)^{k+1} x^k/k = \ln 2$ I.e., $\ln 2$ is the left-hand limit Of $-\ln(1 - x)$ at $x = 1$ .

7.8 Taylor Series Convergence

The Taylor series of $f$ at $a$ is $\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n$ .

Definition. A function $f$ is analytic at $a$ if its Taylor series at $a$ converges to $f(x)$ In some neighborhood of $a$ .

Not every $C^{\infty}$ function is analytic. The standard counterexample is:

$f(x) = \begin{cases} e^{-1/x^2} & x \neq 0 \\ 0 & x = 0 \end{cases}$

$f^{(n)}(0) = 0$ for all $n$ So the Taylor series at $0$ is identically zero, which Converges only to $0$ Not to $f(x)$ for $x \neq 0$ .

7.9 Worked Examples

Worked Example: Show $\sum_{n=1}^{\infty} \frac{x^n}{n^2}$ converges uniformly on $[-1, 1]$

Solution. For $x \in [-1, 1]$ : $\left|\frac{x^n}{n^2}\right| \leq \frac{1}{n^2}$ . Since $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges (it is a $p$ -series with $p = 2 > 1$ ), the Weierstrass M-Test with $M_n = 1/n^2$ implies the series converges uniformly on $[-1, 1]$ . $\blacksquare$

Worked Example: Find the radius of convergence of $\sum_{n=0}^{\infty} \frac{x^n}{n!}$

Solution. Apply the ratio test to the coefficients: $\lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right| = \lim_{n \to \infty} \frac{n!}{(n+1)!} = \lim_{n \to \infty} \frac{1}{n+1} = 0$ .

So $R = \infty$ and the series converges for all $x \in \mathbb{R}$ . This is the power series for $e^x$ . By Theorem 7.4, the derivative of the sum equals $\sum_{n=1}^{\infty} \frac{n x^{n-1}}{n!} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} = \sum_{k=0}^{\infty} \frac{x^k}{k!} = e^x$ , confirming That $e^x$ is its own derivative. $\blacksquare$

Worked Example: Find the radius of convergence of $\sum_{n=1}^{\infty} n! \, x^n$

Solution. Apply the ratio test to the coefficients:

$\lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right| = \lim_{n \to \infty} \frac{(n+1)!}{n!} = \lim_{n \to \infty} (n+1) = \infty$

So $R = 0$ Meaning the series converges only at $x = 0$ . $\blacksquare$

Worked Example: Show $f_n(x) = \frac{x}{1 + nx}$ converges uniformly on $[1, \infty)$

Solution. Pointwise limit: For $x \geq 1$ : $\lim_{n \to \infty} \frac{x}{1 + nx} = \lim_{n \to \infty} \frac{1}{1/x + n} = 0$ .

Uniform convergence: $\sup_{x \in [1, \infty)} \left|\frac{x}{1 + nx} - 0\right| = \sup_{x \geq 1} \frac{x}{1 + nx}$ . To find the maximum, differentiate with respect to $x$ : $\frac{d}{dx}\left(\frac{x}{1+nx}\right) = \frac{1}{(1+nx)^2} > 0$ . So the function is increasing in $x$ on $[1, \infty)$ And:

$\sup_{x \geq 1} \frac{x}{1 + nx} = \lim_{x \to \infty} \frac{x}{1 + nx} = \frac{1}{n}$

Since $\sup |f_n| = 1/n \to 0$ The convergence is uniform on $[1, \infty)$ . $\blacksquare$

:::caution Common Pitfall Pointwise convergence does not preserve continuity, differentiability, or integrability. Uniform Convergence preserves continuity and allows interchange of limit and integral, but not limit and Derivative. For derivatives, uniform convergence of the sequence of derivatives (not the original Sequence) is required, as stated in Theorem 7.4. Also, the Weierstrass M-Test applies only to series Of functions, not sequences; for sequences, one must verify the uniform Cauchy criterion directly.

8. Problem Set

Problem 1. Let $A, B \subseteq \mathbb{R}$ be non-empty and bounded above. Prove that $\sup(A \cup B) = \max(\sup A, \sup B)$ .

Solution

Solution. Let $M = \max(\sup A, \sup B)$ . Without loss, assume $\sup A \geq \sup B$ So $M = \sup A$ . For all $x \in A \cup B$ : either $x \in A$ So $x \leq \sup A = M$ ; or $x \in B$ So $x \leq \sup B \leq M$ . Thus $M$ is an upper bound for $A \cup B$ .

For the least property: since $M = \sup A$ and $A \subseteq A \cup B$ Every upper bound of $A \cup B$ Is an upper bound of $A$ Hence $\geq \sup A = M$ . Therefore $\sup(A \cup B) = M$ . $\blacksquare$

If you get this wrong, revise: Section 1.3 (Supremum and Infimum), Section 1.5 (Properties).

Problem 2. Prove that $\inf A = -\sup(-A)$ for any non-empty bounded set $A \subseteq \mathbb{R}$ .

Solution

Solution. Let $u = \sup(-A)$ . For all $a \in A$ : $-a \in -A$ So $-a \leq u$ Giving $a \geq -u$ . Thus $-u$ is a lower bound for $A$ . If $v$ is any lower bound for $A$ Then $-v$ is an upper bound for $-A$ So $u \leq -v$ I.e., $-u \geq v$ . Hence $-u$ is the greatest lower bound, so $\inf A = -u = -\sup(-A)$ . $\blacksquare$

If you get this wrong, revise: Section 1.5 (Properties of Supremum and Infimum).

Problem 3. Using the $\varepsilon$ - $N$ definition, prove that $\lim_{n \to \infty} \frac{n^2 + 3n}{2n^2 + 1} = \frac{1}{2}$ .

Solution

Solution. Let $\varepsilon > 0$ . Compute:

$\left|\frac{n^2 + 3n}{2n^2 + 1} - \frac{1}{2}\right| = \left|\frac{2(n^2 + 3n) - (2n^2 + 1)}{2(2n^2 + 1)}\right| = \left|\frac{6n - 1}{2(2n^2 + 1)}\right|$

For $n \geq 1$ : $6n - 1 \lt 6n$ and $2n^2 + 1 > 2n^2$ So

$\frac{6n - 1}{2(2n^2 + 1)} \lt \frac{6n}{4n^2} = \frac{3}{2n}$

We need $\frac{3}{2n} \lt \varepsilon$ I.e., $n > 3/(2\varepsilon)$ . Choose $N = \lceil 3/(2\varepsilon) \rceil$ . For $n \geq N$ : the expression is $\lt \varepsilon$ . $\blacksquare$

If you get this wrong, revise: Section 2.1 (Convergence), Section 2.7 (Worked Examples).

Problem 4. Let $a_1 = 1$ and $a_{n+1} = \frac{1}{2}\left(a_n + \frac{2}{a_n}\right)$ . Prove $(a_n)$ converges And find its limit.

Solution

Solution. Step 1: $(a_n)$ is bounded below by $\sqrt{2}$ . By AM-GM: $a_{n+1} = \frac{1}{2}(a_n + 2/a_n) \geq \sqrt{a_n \cdot 2/a_n} = \sqrt{2}$ .

Step 2: $(a_n)$ is decreasing for $n \geq 2$ . Note $a_1 = 1$ , $a_2 = 3/2$ . $a_{n+1} - a_n = \frac{1}{2}(a_n + 2/a_n) - a_n = \frac{1}{2}(2/a_n - a_n) = \frac{2 - a_n^2}{2a_n}$ . Since $a_n \geq \sqrt{2}$ for $n \geq 2$ , $a_n^2 \geq 2$ So $a_{n+1} - a_n \leq 0$ .

Step 3: By the Monotone Convergence Theorem, $L = \lim a_n$ exists. Taking limits: $L = \frac{1}{2}(L + 2/L)$ Giving $2L = L + 2/L$ So $L = 2/L$ Hence $L^2 = 2$ . Since $a_n \geq \sqrt{2}$ for $n \geq 2$ , $L \geq 0$ So $L = \sqrt{2}$ . $\blacksquare$

If you get this wrong, revise: Section 2.2 (Monotone Convergence Theorem), Section 2.7 (recursive sequences).

Problem 5. Compute $\limsup_{n \to \infty} a_n$ and $\liminf_{n \to \infty} a_n$ for $a_n = 2 + (-1)^n \frac{n}{n+1}$ .

Solution

Solution. Write $a_n = 2 + (-1)^n \cdot \frac{n}{n+1}$ .

For even $n = 2k$ : $a_{2k} = 2 + \frac{2k}{2k+1} \to 2 + 1 = 3$ . For odd $n = 2k - 1$ : $a_{2k-1} = 2 - \frac{2k-1}{2k} \to 2 - 1 = 1$ .

Since these are the only two subsequential limits: $\limsup a_n = 3$ and $\liminf a_n = 1$ . The sequence diverges since $\limsup \neq \liminf$ . $\blacksquare$

If you get this wrong, revise: Section 2.6 (Limit Superior and Limit Inferior).

Problem 6. Determine whether $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2}$ converges.

Solution

Solution. Apply the integral test with $f(x) = 1/(x(\ln x)^2)$ on $[2, \infty)$ . The function is Positive, continuous, and decreasing. Compute via $u = \ln x$ , $du = dx/x$ :

$\int_2^{\infty} \frac{1}{x(\ln x)^2}\, dx = \int_{\ln 2}^{\infty} \frac{1}{u^2}\, du = \left[-\frac{1}{u}\right]_{\ln 2}^{\infty} = \frac{1}{\ln 2} \lt \infty$

The integral converges, so by the integral test, the series converges. $\blacksquare$

If you get this wrong, revise: Section 3.2 (Integral Test), Section 3.6 (Worked Examples).

Problem 7. Does $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{1/3}}$ converge absolutely, conditionally, or diverge?

Solution

Solution. The absolute series is $\sum 1/n^{1/3}$ Which is a $p$ -series with $p = 1/3 \lt 1$ So it diverges. Hence the series does not converge absolutely.

For conditional convergence, apply the alternating series test: $a_n = 1/n^{1/3}$ is positive, Decreasing, and $a_n \to 0$ . Therefore $\sum (-1)^{n+1}/n^{1/3}$ converges.

Since it converges but not absolutely, it converges conditionally. $\blacksquare$

If you get this wrong, revise: Section 3.3 (Absolute and Conditional Convergence), Section 3.6 (Alternating Series Test).

Problem 8. Find the sum of $\sum_{n=1}^{\infty} \frac{1}{n(n+2)}$ .

Solution

Solution. Use partial fractions: $\frac{1}{n(n+2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+2}\right)$ . The $N$ -th partial sum telescopes:

$S_N = \frac{1}{2}\left[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{N} - \frac{1}{N+2}\right)\right]$

Most terms cancel. The surviving terms are:

$S_N = \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{N+1} - \frac{1}{N+2}\right)$

As $N \to \infty$ : $S_N \to \frac{1}{2}(1 + 1/2) = \frac{3}{4}$ . $\blacksquare$

If you get this wrong, revise: Section 3.1 (Definitions and Convergence), telescoping series.

Problem 9. Give an explicit rearrangement of $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ whose sum is $0$ .

Solution

Solution. By the Riemann Rearrangement Theorem, such a rearrangement exists. We construct it Explicitly. The positive terms are $1, 1/3, 1/5, \ldots$ and the negative terms are $-1/2, -1/4, -1/6, \ldots$ .

Start: $S_1 = 1$ . Then add negative terms until we go below $0$ : $S_2 = 1 - 1/2 = 1/2 > 0$ . $S_3 = 1 - 1/2 - 1/4 = 1/4 > 0$ . $S_4 = 1 - 1/2 - 1/4 - 1/6 = -1/12 \lt 0$ .

Then add positive terms until we exceed $0$ : $S_5 = -1/12 + 1/3 = 1/4 > 0$ .

Then add negative terms until below $0$ : $S_6 = 1/4 - 1/8 = 1/8 > 0$ . $S_7 = 1/8 - 1/10 = 1/40 > 0$ . $S_8 = 1/40 - 1/12 = -7/120 \lt 0$ .

Continue this process. Since $\sum 1/(2k-1) = \infty$ and $\sum 1/(2k) = \infty$ We can always Continue. Since $1/n \to 0$ The oscillations shrink to $0$ . The resulting rearrangement converges to $0$ . $\blacksquare$

If you get this wrong, revise: Section 3.5 (Rearrangement of Series).

Problem 10. Prove using $\varepsilon$ - $\delta$ that $f(x) = x^3$ is continuous at every $a \in \mathbb{R}$ .

Solution

Solution. Let $a \in \mathbb{R}$ and $\varepsilon > 0$ . Compute:

$|f(x) - f(a)| = |x^3 - a^3| = |x - a| \cdot |x^2 + ax + a^2|$

Restrict to $|x - a| \lt 1$ So $|x| \lt |a| + 1$ Giving $|x^2 + ax + a^2| \leq (|a|+1)^2 + |a|(|a|+1) + a^2 = 3a^2 + 3|a| + 1$ . Let $M = 3a^2 + 3|a| + 1$ .

Choose $\delta = \min(1, \varepsilon/M)$ . Then $|x - a| \lt \delta$ implies:

$|x^3 - a^3| \leq |x - a| \cdot M \lt \frac{\varepsilon}{M} \cdot M = \varepsilon$

$\blacksquare$

If you get this wrong, revise: Section 4.2 (Continuity), Section 4.7 (Worked Examples).

Problem 11. Prove that $f(x) = x \sin(1/x)$ (with $f(0) = 0$ ) is continuous on $\mathbb{R}$ but not Uniformly continuous on $(0, 1)$ . (Trick question --- see solution.)

Solution

Solution. Continuity at $0$ : Given $\varepsilon > 0$ Choose $\delta = \varepsilon$ . For $|x - 0| = |x| \lt \delta$ : $|f(x) - f(0)| = |x \sin(1/x)| \leq |x| \lt \delta = \varepsilon$ . So $f$ is continuous at $0$ . For $x \neq 0$ , $f$ is a product of continuous functions, hence continuous.

On uniform continuity: Actually, $f(x) = x\sin(1/x)$ is uniformly continuous on $(0, 1)$ ! Here is why: $f$ extends continuously to $[0, 1]$ (define $f(0) = 0$ ). By the Heine-Cantor theorem (Theorem 4.5), $f$ is uniformly continuous on $[0, 1]$ And hence on the subset $(0, 1)$ .

The function that is not uniformly continuous on $(0, 1)$ is $g(x) = \sin(1/x)$ Which does not Extend continuously to $0$ . Or $h(x) = 1/x$ Which is unbounded. But $f(x) = x\sin(1/x)$ is bounded And has a continuous extension, so it is uniformly continuous. $\blacksquare$

If you get this wrong, revise: Section 4.5 (Uniform Continuity), Section 4.6 (Heine-Cantor).

Problem 12. Prove that if $f'(x) = g'(x)$ for all $x \in (a, b)$ Then $f(x) = g(x) + C$ for some Constant $C$ .

Solution

Solution. Let $h(x) = f(x) - g(x)$ . Then $h'(x) = f'(x) - g'(x) = 0$ for all $x \in (a, b)$ . By Corollary 5.4 (a consequence of the Mean Value Theorem), $h$ is constant on $(a, b)$ . So $f(x) - g(x) = C$ for some $C \in \mathbb{R}$ I.e., $f(x) = g(x) + C$ . $\blacksquare$

If you get this wrong, revise: Section 5.3 (Mean Value Theorem, Corollary 5.4).

Problem 13. Use Taylor’s theorem with remainder to bound the error in approximating $e^{0.2}$ Using the fourth-degree Maclaurin polynomial.

Solution

Solution. The fourth-degree Maclaurin polynomial of $e^x$ is:

$T_4(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}$

By Taylor’s theorem, $R_4(x) = \frac{e^{\xi}}{5!} x^5$ for some $\xi$ between $0$ and $x$ . For $x = 0.2$ : $\xi \in (0, 0.2)$ So $e^{\xi} \lt e^{0.2} \lt e^{1/4} \lt 1.3$ .

$|R_4(0.2)| = \frac{e^{\xi}}{120} (0.2)^5 \lt \frac{1.3}{120} \cdot 0.00032 = \frac{1.3 \times 0.00032}{120} \approx 3.47 \times 10^{-6}$

So $T_4(0.2) = 1 + 0.2 + 0.02 + 0.001333 + 0.000067 = 1.221400$ approximates $e^{0.2}$ with Error less than $3.5 \times 10^{-6}$ . $\blacksquare$

If you get this wrong, revise: Section 5.4 (Taylor’s Theorem), Section 5.7 (Worked Examples).

Problem 14. Compute $\int_0^1 x^3\, dx$ from the definition using upper and lower Riemann sums.

Solution

Solution. Let $P_n = \{0, 1/n, 2/n, \ldots, 1\}$ . On $[(i-1)/n, i/n]$ , $f(x) = x^3$ has $M_i = (i/n)^3$ and $m_i = ((i-1)/n)^3$ .

$U(f, P_n) = \sum_{i=1}^{n} \frac{i^3}{n^3} \cdot \frac{1}{n} = \frac{1}{n^4} \sum_{i=1}^{n} i^3 = \frac{1}{n^4} \cdot \frac{n^2(n+1)^2}{4}$

As $n \to \infty$ :

$\lim_{n \to \infty} U(f, P_n) = \lim_{n \to \infty} \frac{(n+1)^2}{4n^2} = \frac{1}{4}$

Similarly, $L(f, P_n) \to 1/4$ . So $\int_0^1 x^3\, dx = 1/4$ . $\blacksquare$

If you get this wrong, revise: Section 6.1 (Definition), Section 6.5 (Worked Examples).

Problem 14b. Show that the Dirichlet function $f(x) = \begin{cases} 1 & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}$ is not Riemann integrable on $[0, 1]$ .

Solution

Solution. Every non-empty subinterval $[x_{i-1}, x_i]$ of any partition contains both rational and Irrational numbers (by density of $\mathbb{Q}$ and density of $\mathbb{R} \setminus \mathbb{Q}$ ). So $M_i = \sup f = 1$ and $m_i = \inf f = 0$ for every subinterval.

For any partition $P$ : $U(f, P) = \sum 1 \cdot \Delta x_i = 1$ and $L(f, P) = \sum 0 \cdot \Delta x_i = 0$ . Hence $\overline{\int_0^1} f = 1 \neq 0 = \underline{\int_0^1} f$ So $f$ is not Riemann integrable.

This also follows from Lebesgue’s criterion: $f$ is discontinuous everywhere, and $[0,1]$ does not Have measure zero. $\blacksquare$

If you get this wrong, revise: Section 6.2 (Integrability Criteria), Theorem 6.4b.

Problem 15. Evaluate $\int_0^1 \frac{x}{\sqrt{1 - x^2}}\, dx$ as an improper integral.

Solution

Solution. The integrand $f(x) = x/\sqrt{1 - x^2}$ is unbounded as $x \to 1^-$ . This is a Type II Improper integral.

$\int_0^1 \frac{x}{\sqrt{1 - x^2}}\, dx = \lim_{b \to 1^-} \int_0^b \frac{x}{\sqrt{1 - x^2}}\, dx$

Compute via substitution $u = 1 - x^2$ , $du = -2x\, dx$ :

$= \lim_{b \to 1^-} \left[-\sqrt{1 - x^2}\right]_0^b = \lim_{b \to 1^-} \left(-\sqrt{1 - b^2} + 1\right) = 0 + 1 = 1$

The improper integral converges to $1$ . $\blacksquare$

If you get this wrong, revise: Section 6.6 (Improper Integrals).

Problem 16. Let $f_n(x) = \frac{nx}{1 + n^2 x^2}$ on $(0, \infty)$ . Find the pointwise limit and Determine whether the convergence is uniform on $(0, \infty)$ .

Solution

Solution. Pointwise limit: For fixed $x > 0$ : $\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{nx}{1 + n^2 x^2} = \lim_{n \to \infty} \frac{x/n}{1/n^2 + x^2} = 0$ .

So $f_n \to 0$ pointwise on $(0, \infty)$ .

Uniform convergence? We check $\sup_{x > 0} |f_n(x) - 0| = \sup_{x > 0} \frac{nx}{1 + n^2 x^2}$ . To maximize, differentiate with respect to $x$ (treating $n$ as fixed):

$\frac{d}{dx}\left(\frac{nx}{1 + n^2 x^2}\right) = \frac{n(1 + n^2 x^2) - nx \cdot 2n^2 x}{(1 + n^2 x^2)^2} = \frac{n - n^3 x^2}{(1 + n^2 x^2)^2}$

Setting to zero: $n - n^3 x^2 = 0$ So $x = 1/n$ . The maximum value is $f_n(1/n) = \frac{n \cdot 1/n}{1 + n^2/n^2} = \frac{1}{2}$ .

Since $\sup_{x > 0} |f_n(x)| = 1/2$ for all $n$ This does not tend to $0$ . Therefore the convergence Is not uniform on $(0, \infty)$ . $\blacksquare$

If you get this wrong, revise: Section 7.2 (Uniform Convergence), Section 7.1 (Pointwise Convergence).

Problem 17. Find the radius of convergence of $\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2} x^n$ .

Solution

Solution. Apply the ratio test to the terms:

$\left|\frac{a_{n+1}}{a_n}\right| = \frac{(2(n+1))!}{((n+1)!)^2} \cdot \frac{(n!)^2}{(2n)!} \cdot |x| = \frac{(2n+2)(2n+1)}{(n+1)^2} \cdot |x|$

$= \frac{2(2n+1)}{n+1} \cdot |x| = \frac{4n + 2}{n + 1} \cdot |x| \to 4|x| \quad \mathrm{as\ } n \to \infty$

The series converges when $4|x| \lt 1$ I.e., $|x| \lt 1/4$ And diverges when $4|x| > 1$ . The radius of convergence is $R = 1/4$ . $\blacksquare$

If you get this wrong, revise: Section 7.7 (Power Series), Section 3.2 (Ratio Test).

Problem 18. Let $f_n(x) = x^n/n$ on $[0, 1]$ . Show that $f_n \to 0$ uniformly, but $f_n'(x) = x^{n-1}$ does not converge uniformly on $(0, 1)$ .

Solution

Solution. Uniform convergence of $f_n$ : $\sup_{x \in [0,1]} |x^n/n| = 1/n \to 0$ as $n \to \infty$ . So $f_n \to 0$ uniformly on $[0, 1]$ .

Non-uniform convergence of $f_n'$ : $f_n'(x) = x^{n-1}$ . The pointwise limit is $g(x) = 0$ for $0 \leq x \lt 1$ and $g(1) = 1$ . So $\sup_{x \in [0,1]} |f_n'(x) - g(x)| \geq |f_n'(1) - g(1)| = |1 - 1| = 0$ .

Actually, check $\sup_{x \in [0,1)} |x^{n-1}| = 1$ (approached as $x \to 1^-$ ). But $g(x) = 0$ on $[0, 1)$ So $\sup_{x \in [0,1)} |x^{n-1} - 0| = 1$ for all $n$ . This does not Tend to $0$ So $f_n'$ does not converge uniformly on $[0, 1)$ .

This illustrates that uniform convergence of functions does not imply uniform convergence of Derivatives, which is why Theorem 7.4 requires the stronger hypothesis of uniform convergence of $(f_n')$ . $\blacksquare$

If you get this wrong, revise: Section 7.2 (Uniform Convergence), Section 7.6 (Uniform Convergence and Differentiation).

Problem 19. Let $f_n(x) = \frac{x}{1 + nx^2}$ on $[0, \infty)$ . Find the pointwise limit and determine Whether the convergence is uniform.

Solution

Solution. Pointwise limit: For $x = 0$ : $f_n(0) = 0$ for all $n$ . For $x > 0$ : $\lim_{n \to \infty} \frac{x}{1 + nx^2} = \lim_{n \to \infty} \frac{1}{1/x + nx} = 0$ . So $f_n \to 0$ pointwise.

Uniform convergence on $[0, \infty)$ ? We check $\sup_{x \geq 0} |f_n(x)|$ . Differentiate: $\frac{d}{dx}\left(\frac{x}{1 + nx^2}\right) = \frac{1 - nx^2}{(1 + nx^2)^2}$ . Setting to zero: $x = 1/\sqrt{n}$ . The maximum value is $f_n(1/\sqrt{n}) = \frac{1/\sqrt{n}}{1 + n/n} = \frac{1}{2\sqrt{n}}$ .

Since $\sup_{x \geq 0} |f_n(x)| = \frac{1}{2\sqrt{n}} \to 0$ as $n \to \infty$ The convergence is Uniform on $[0, \infty)$ . $\blacksquare$

If you get this wrong, revise: Section 7.2 (Uniform Convergence).

Problem 20. Prove that the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1}$ converges uniformly on $[-1, 1]$ and identify its sum.

Solution

Solution. For $|x| \leq 1$ : $\left|\frac{(-1)^n x^{2n+1}}{2n+1}\right| \leq \frac{1}{2n+1}$ . The series $\sum \frac{1}{2n+1}$ diverges (it dominates half the harmonic series), so the Weierstrass M-test does not apply directly with these bounds.

However, by the alternating series test, the series converges pointwise for every $|x| \leq 1$ (since $\frac{|x|^{2n+1}}{2n+1}$ decreases to $0$ for $|x| \leq 1$ ). The sum is $\arctan x$ Which is the Taylor series of $\arctan$ about $0$ .

For uniform convergence, we use Abel’s test for uniform convergence of series: if $\sum f_n(x)$ has uniformly bounded partial sums and $g_n(x)$ decreases uniformly to $0$ Then $\sum f_n(x) g_n(x)$ converges uniformly. Here $f_n(x) = (-1)^n x^{2n+1}$ and $g_n(x) = 1/(2n+1)$ Is independent of $x$ .

The partial sums $\left|\sum_{k=0}^{n} (-1)^k x^{2k+1}\right| \leq \frac{|x|}{1 + x^2} \leq \frac{1}{2}$ for $|x| \leq 1$ (geometric series bound). And $1/(2n+1) \to 0$ uniformly. By Abel’s test, the Convergence is uniform on $[-1, 1]$ .

Setting $x = 1$ : $\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \arctan 1 = \pi/4$ . $\blacksquare$

If you get this wrong, revise: Section 7.3 (Weierstrass M-Test), Section 7.7 (Power Series), Abel’s theorem.

Common Pitfalls

Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).
Misreading the question, particularly with ‘hence’ vs ‘hence or otherwise’. The former requires using previous work.
Confusing pointwise and uniform convergence. Pointwise convergence does not guarantee uniform convergence.

Worked Examples

Example 1: Epsilon-Delta Proof of Continuity

Problem. Prove that $f(x) = 3x + 1$ is continuous at $x = 2$ using the $\varepsilon$ - $\delta$ definition.

Solution. Let $\varepsilon > 0$ . Choose $\delta = \varepsilon / 3$ .

If $0 < |x - 2| < \delta$ , then:

$|f(x) - f(2)| = |3x + 1 - 7| = 3|x - 2| < 3 \cdot \frac{\varepsilon}{3} = \varepsilon$

Since for every $\varepsilon > 0$ we found $\delta > 0$ satisfying the condition, $f$ is continuous at $x = 2$ .

$\blacksquare$

Example 2: Convergence of a Sequence

Problem. Prove that $a_n = \frac{2n+1}{n+3}$ converges and find its limit.

Solution. Claim: $a_n \to 2$ .

$|a_n - 2| = \left|\frac{2n+1}{n+3} - 2\right| = \left|\frac{2n+1 - 2n - 6}{n+3}\right| = \frac{5}{n+3}$ .

For any $\varepsilon > 0$ , choose $N > \frac{5}{\varepsilon} - 3$ .

For $n > N$ : $|a_n - 2| = \frac{5}{n+3} < \frac{5}{N+3} < \varepsilon$ .

Therefore $a_n \to 2$ by definition.

$\blacksquare$

Summary

Completeness of $\mathbb{R}$ : every non-empty bounded-above set has a supremum; equivalent to the monotone convergence theorem and Bolzano-Weierstrass.
Limits and continuity: $\varepsilon$ - $\delta$ definition; sequential characterisation; intermediate value theorem and extreme value theorem.
Differentiability: $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ ; mean value theorem; Taylor’s theorem with remainder.
Riemann integration: defined via upper and lower sums; fundamental theorem of calculus connects integration and differentiation.
Sequences and series: tests for convergence (comparison, ratio, root, integral); absolute vs conditional convergence; power series and radius of convergence.

Cross-References

Topic	Site	Link
Complex Analysis	WyattsNotes	View
Linear Algebra	WyattsNotes	View
Multivariable Calculus	WyattsNotes	View
Real Analysis — MIT 18.100	MIT OCW	View

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Real Analysis

1. The Real Number System

1.1 Field Axioms

1.2 Order and the Completeness Axiom

1.3 Supremum and Infimum

1.4 Archimedean Property

1.5 Properties of Supremum and Infimum

1.6 Construction of R\mathbb{R}R via Dedekind Cuts

1.7 Equivalences of Completeness

2. Sequences and Limits

2.1 Convergence

2.2 Convergence Theorems

2.3 Cauchy Sequences

2.4 Subsequences

2.5 The Bolzano-Weierstrass Theorem

2.6 Limit Superior and Limit Inferior

2.7 Worked Examples

3. Series

3.1 Definitions and Convergence

3.2 Convergence Tests

3.3 Absolute and Conditional Convergence

3.4 The Alternating Series Estimation Theorem

3.5 Cauchy Condensation Test

3.6 Rearrangement of Series

3.7 Worked Examples

4. Continuity

4.1 Limits of Functions

4.2 Continuity

4.2a Sequential Characterization of Limits and Continuity

4.3 Intermediate Value Theorem

4.4 Extreme Value Theorem

4.5 Uniform Continuity

4.6 The Heine-Cantor Theorem

4.7 Worked Examples

5. Differentiability

5.1 The Derivative

5.2 Differentiation Rules

5.3 Mean Value Theorem

5.4 Taylor’s Theorem

5.5 L’Hôpital’s Rule

5.6 Darboux’s Theorem

5.7 Worked Examples

6. Riemann Integration

6.1 Definition

6.2 Integrability Criteria

6.3 Properties of the Integral

6.4 The Fundamental Theorem of Calculus

6.5 Worked Examples

6.6 Improper Integrals

7. Sequences and Series of Functions

7.1 Pointwise Convergence

7.2 Uniform Convergence

7.3 The Weierstrass M-Test

7.4 Uniform Convergence and Continuity

7.5 Uniform Convergence and Integration

7.6 Uniform Convergence and Differentiation

7.7 Power Series

7.8 Taylor Series Convergence

7.9 Worked Examples

8. Problem Set

Common Pitfalls

Worked Examples

Example 1: Epsilon-Delta Proof of Continuity

Example 2: Convergence of a Sequence

Summary

Cross-References

1.6 Construction of $\mathbb{R}$ via Dedekind Cuts