Definition – Limit of a sequence

Contents
1. Definition

(!) Convergence and divergence both direct to this page.

Definition

Let $s_n$ be a sequence. If there exists an $L\in{\mathbb{R}}$ such that

$\forall \epsilon >0,~ \exists N>0 \text{~s.t~} n>N \implies |s_n - L| < \epsilon$ .

Then we write $\lim_{n \rightarrow \infty} s_n = L$ , and we say that $s_n$ converges. If no such $L$ exists, then we say that $s_n$ diverges.

Theorem – Every convergent sequence is bounded.

Proof – Let $(s_n)$ be a sequence. We have from the definition of convergence that for every $\epsilon > 0$ , there exists an $N>0$ such that

$n>N \implies |s_n - L| < \epsilon\quad \bf{(1)}$

Where $L$ is the limit of $(s_n)$ . Now, $\bf{(1)}$ can be rewritten as

$n>N \implies -\epsilon < s_n - L <\epsilon \implies -L - \epsilon < s_n < L + \epsilon$

However, this gives two inequalities, firstly $-L < s_n + \epsilon \implies -L \leq s_n$ , and $s_n < L + \epsilon \implies s_n \leq L$ . These two inequalities prove that $(s_n)$ is bounded above/below by $-L$ and $L$ respectively.

Theorem – If $(s_n)$ is a convergent sequence with limit $L$ , then every subsequence of $(s_n)$ converges to $L$ .

Theorem – Every convergent sequence, $(s_n)$ , has a convergent subsequence.

Proof – We have two separate proofs based on whether $(s_n)$ is finite or infinite. If $(s_n)$ is finite, then we have that there is an infinite number of $i\in\mathbb{N}$ such that $s_i = x$ for some $x$ . Thus, $(s_n)$ converges to $x$ because for all $n>N_i$ , $|s_n - L| = 0 < \epsilon$ .

Now, if $(s_n)$ is infinite, since it is bounded by assumption, there exists at least one accumulation point, $x$ (often referred to as a limit point) of $(s_n)$ . But then, we may construct consecutive neighborhoods $N(x,\epsilon)$ . But then $(s_n) \rightarrow L$ .

Theorem – Every unbounded sequence, $(s_n)$ , contains a monotone subsequence that has either $\pm\infty$ as its limit.

Proof – We already know that an unbounded increasing sequence has limit $\infty$ , and that an unbounded decreasing sequence has limit $-\infty$ , so it just remains to construct such a subsequence from $(s_n)$ . We may do so as a consequence of the hypothesis. We may easily choose one element to be in the subsequence. Then, assume we already have taken $k$ elements in the subsequence, $(s_{n_1}, ..., s_{n_k})$ . Since $(s_n)$ is infinite and unbounded, it follows that there exists another element, and if it were the highest element in $(s_n)$ , then $(s_n)$ would be bounded. It follows that another element $s_{n_{k+1}}$ exists such that $s_{n_k} \leq s_{n_{k+1}}$ . But this completes the proof, as we have shown that such a subsequence may be constructed.

A similar method may be used if we assume that $(s_n)$ is unbounded below.