# $\{a_n\}$ is a Cauchy sequence, if $a_{n+2} = \frac{a_n + a_{n+1}}{2}$

Suppose that the sequence $\{a_n\}$ satisies the relation $$a_{n+2} = \frac{a_n + a_{n+1}}{2},$$ for all $n \in \mathbb{N}_{+}$

Prove that $\{a_n\}$ is a Cauchy sequence

Note that $$a_{n+2}-a_{n+1}=-\frac{a_{n+1}-a_n}{2}=\cdots=(-1)^{n}\frac{a_2-a_1}{2^n},$$ and hence $$\sum_{n=1}^\infty \lvert a_{n+1}-a_n\rvert = 2\lvert a_2-a_1\rvert <\infty.$$ In particular, for $m\ge n$, it is not hard to show that $$\lvert a_m-a_n\rvert\le 2^{-n+1}\lvert a_2-a_1\rvert.$$ Finally, $$\lim_{n\to\infty}a_n= \frac{a_1+2a_2}{3},$$ as $$a_n=\frac{a_1+2a_2}{3}-\frac{a_1-a_2}{3\cdot(-2)^{n-2}}.$$

$$a_{n+2}-a_{n+1}=\frac{a_n-a_{n+1}}{2}$$

and this gives

$$a_{n+2}-a_{n+1}=\frac{a_{1}-a_{0}}{2^n}$$

So if we look at

$$|a_m-a_n|\leq |a_{n}-a_{n+1}|+|a_{n+1}-a_{n+2}|\cdots +|a_{m-1}-a_m| \leq |a_1-a_0|\left(\frac{1}{2^n}+\dots +\frac{1}{2^m}\right)$$ we see that by choosing $n$ and $m$ large enough the difference will be smaller than $\epsilon$ showing that the sequence is cauchy.

Let $d_n=|a_{n+1}-a_n|$. Show $d_{n+1}=\frac12 d_n$. Conclude that $|a_n-a_m|\le 2^{-N}|a_1-a_0|$ if $n,m>N$.