Prove: if $s_n$ is Cauchy Sequence, then $s_n$ converges as $n \to \infty$

Question

How to prove the following:

if $s_n$ is Cauchy Sequence, then $s_n$ converges as $n \to \infty$

Using the following theorem:

A Cauchy Sequence is bounded.

Assumption: 1. $s_n$ is sequence of real numbers.

Answer 1

Let $\{S_n\}$ be Cauchy Sequence of real numbers.

Every Cauchy Sequence is bounded.

$\implies$ $\{S_n\}$ is bdd seq, any bdd squ has convergent sub seq.

i.e $\{S_n\}$ has convergent sub seq.

say$\{S_{n_k}\}$

Let $lim_{k \to \infty} S_{n_k}=L$

using the above fact show that

$lim_{k \to \infty} S_n=L$

$|S_n-L|=|S_n-S_{n_k}+S_{n_k}-L|\leq |S_n-S_{n_k}|+|S_{n_k}-L|$

fix $\epsilon>0 $. Since $\{S_n\}$ is Cauchy sequence .Therefore for $\epsilon>0 $, there exist integer $N_1$such that $$|S_n-S_m|<\frac{\epsilon}{2} ,\forall n,m \geq N_1$$ Since $lim_{k \to \infty} S_{n_k}=L$. For $\epsilon>0 $

there exist integer $N_2$such that $$|S_{n_k}-L|<\frac{\epsilon}{2} ,\forall n_k \geq N_2$$

Now choose N=max$\{N_1,N_2 \}$.So that

$|S_n-L|\leq |S_n-S_{n_k}|+|S_{n_k}-L|<\frac{\epsilon}{2}+\frac{\epsilon}{2}$

$\implies |S_n-L|<\epsilon \forall n\geq N$

which means $lim_{k \to \infty} S_n=L$.

Answer 2

Hint: Since $s_n$ is bounded, it has a convergent subsequence $s_{n_k}$ converging to $L$. Show that $s_n$ converges to $L$ by estimating $|s_n - L|$ using the Cauchy sequence condition.

Answer 3

This is an analysis rite of passage, so you should work out the details, but here is a hint:

Let $a_n$ be a cauchy sequence. Then it is bounded (can you show this?)

Use Bolzano-Weierstrass, to take a monotone subsequence. then there is a convergent subsequence, call its limit $L$. what does this mean for $a_n$? You will probably need the triangle inequality!

Answer 4

Before reading this post in its entirety, please just read the statements of the lemmas and see if you can prove them yourself.

Lemma $1$: A bounded monotonic (nonincreasing or nondecreasing) sequence converges.

Proof: Suppose $(a_n)$ is a bounded monotonic sequence. Without loss of generality, suppose it is nondecreasing. Then let $a = \sup_{n\in\Bbb{N}} a_n$. We claim that $a$ is the limit of $(a_n)$. To show this, let $\epsilon>0$. A basic property of suprema states that if $s = \sup S$, then for all $\epsilon>0$ there exists $t\in S$ so that $|t-s|<\epsilon$. Therefore, let $N\in\Bbb{N}$ so that $|a_N-a|<\epsilon$. Then for any $m>N$, we have $|a-a_N| = |a-a_m + a_m - a_N| = |a-a_m| + |a_m-a_N| < \epsilon$ (since both terms are nonnegative), and so $|a-a_m| < \epsilon$. This proves the lemma.

Lemma $2$: A bounded sequence contains a monotonic subsequence.

Proof: Suppose $(a_n)$ is bounded. We analyze two cases.

First, if possible, define $b_n = \max_{k\geq n} a_k$, deleting any repeating terms. Then clearly $(b_n)$ is nonincreasing and bounded.

Now suppose it is impossible to construct such $(b_n)$; this means that for some $N\in\Bbb{N}$, $m>N$ implies that there exists $m'>m$ so that $a_{m'} > a_m$ (since there is "no maximal" term after $N$). Then let $c_1 = a_N$, and for $n>1$ choose $c_n = a_k > c_{n-1}$, since such $a_k$ is guaranteed to exist. Then $(c_n)$ is bounded and monotonic (in fact, strictly increasing).

Lemma $3$: If a subsequence of a Cauchy sequence converges, then the Cauchy sequence converges to the same value.

Proof: Let $(a_n)$ be a Cauchy sequence, and $a_{n_i}\to a$ a convergent subsequence. Pick $\epsilon > 0$. Then let $N \in \Bbb{N}$ so that $m,p>N\implies |a_p-a_m| < \epsilon/2$, and such that $m>N$ implies $|a-a_{n_m}|\leq \epsilon/2$. Note that $n_i\geq i$. Then if $m>N$, then $|a-a_m| = |a-a_{n_m}+a_{n_m}-a_m|\leq |a-a_{n_m}|+|a_{n_m}-a_m| < \epsilon/2 + \epsilon/2 = \epsilon$. So $a_n\to a$.

These three lemmas, along with the fact that every Cauchy sequence is bounded, imply that every Cauchy sequence converges.