Sequences and Limits

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 $n$. $\blacksquare$

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 $n \geq N$. $\blacksquare$

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

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

$\blacksquare$

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.

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 $a_{n_k} \to c$. $\blacksquare$

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 $\limsup a_n = \liminf a_n$.

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 $\limsup = 0$.

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