23 - Cauchy Sequence

Cauchy Sequence

Let $(a_n)$ be a sequence then
$(a_n)$ is a Cauchy Sequence if

$$\forall ϵ>0\exists N\in \mathbb{N}\text{ such that }\forall m,n\ge N:|a_n-a_m|<ϵ$$

Convergent Sequences are Cauchy

Let $(a_n)$ be a convergent sequence then

$$(a_n)\text{ is Cauchy}$$

Proof

Say $a_n\to L\text{ as }n\to \infty$
Take $ϵ>0$
Since $a_n\to L$, there is $N$ such that

$$\text{if }n\ge N\text{ then }|a_n-L|<\frac{ϵ}{2}$$

For $m,n\ge N$ then

$$|a_m-L|<\frac{ϵ}{2}\text{ and }|a_n-L|<\frac{ϵ}{2}$$

So by Triangle Inequality

$$\begin{array}{rl}|a_m-a_n|& =|(a_m-L)+(L-a_n)|\\ & \le |a_m-L|+|a_n-L|<ϵ\end{array}$$

Hence $(a_n)$ is Cauchy

Cauchy sequences are bounded

Let $(a_n)$ be a Cauchy sequence then

$$(a_n)\text{ is bounded}$$

Proof

Since $(a_n)$ is Cauchy then using $ϵ=1$ then there is $N$ such that

$$\text{if }m,n\ge N\text{ then }|a_m-a_n|<1$$

Now for $n\ge N$ we have

$$|a_n-a_N|<1$$

So

$$|a_n|=|(a_n-a_N)+a_N|\le 1+|a_N|$$

Then let

$$K=\max \{|a_1|,|a_2|,\cdots ,|a_{N-1}|,1+|a_N|\}$$

Then

$$|a_n|\le K\text{ for all }n\ge 1$$

Hence $(a_n)$ is bounded

Convergence of a subsequence of Cauchy Sequences

Let $(a_n)$ be a Cauchy sequence
Suppose that the subsequence $(a_{n_r}{)}_r$ converges then

$$(a_n)\text{ converges}$$

Proof

Say that $a_{n_r}\to L$ as $r\to \infty$
For $ϵ>0$
There is $N_1$ such that if $r\ge N_1$ then

$$|a_{n_r}-L|<\frac{ϵ}{2}$$

Since $(a_n)$ is Cauchy there is $N_2$ such that

$$\text{if }m,n\ge N_2\text{ then }|a_m-a_n|<\frac{ϵ}{2}$$

Let $N=\max \{N_1,N_2\}$
Let $r=N$ then $n_r\ge r\ge N_1$ so $|a_{n_r}-L|<\frac{ϵ}{2}$

If $n\ge N$ then $n,n_r\ge N_2$ so $|a_{n_r}-a_n|<\frac{ϵ}{2}$
So

$$\begin{array}{rl}|a_n-L|& =|(a_n-a_{n_r})+(a_{n_r}-L)|\\ & \le |a_n-a_{n_r}|+|a_{n_r}-L|<ϵ\end{array}$$

So $a_n\to L$ as $n\to \infty$