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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

I've self-taught myself sequences, so please provide a detailed answer.

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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 $$ a_n=\frac{a_1+2a_2}{3}-\frac{a_1-a_2}{3\cdot(-2)^{n-2}}. $$

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

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.

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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 $n,m>N$.

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