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:

1. Reflexivity: $a \leq a$
2. Antisymmetry: $a \leq b$ and $b \leq a$ implies $a = b$
3. Transitivity: $a \leq b$ and $b \leq c$ implies $a \leq c$
4. Totality: for all $a, b$Either $a \leq b$ or $b \leq a$
5. Compatibility with addition: $a \leq b$ implies $a + c \leq b + c$
6. 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:

1. $u$ is an upper bound: $s \leq u$ for all $s \in S$.
2. $u$ 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 $-u = \inf(S)$. $\blacksquare$

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:

1. $\alpha \neq \emptyset$ and $\alpha \neq \mathbb{Q}$
2. If $p \in \alpha$ and $q \lt p$ (with $q \in \mathbb{Q}$), then $q \in \alpha$ (downward closure)
3. $\alpha$ 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:

1. Least Upper Bound Property: Every non-empty set bounded above has a supremum.
2. Monotone Convergence Theorem: Every bounded monotone sequence converges.
3. 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 $\bigcap I_n$.
4. Bolzano-Weierstrass Property: Every bounded sequence has a convergent subsequence.
5. 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:

1. $a_n + b_n \to L + M$
2. $a_n b_n \to LM$
3. $a_n / b_n \to L/M$ (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:

1. $\limsup(a_n + b_n) \leq \limsup a_n + \limsup b_n$
2. $\liminf(a_n + b_n) \geq \liminf a_n + \liminf b_n$
3. 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 $\limsup a_n = \liminf a_n$.

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 $\frac{1}{n+1} \lt \varepsilon$. $\blacksquare$

:::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:

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