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This question already has an answer here:

Let's suppose that $a_n$ satisfy $|a_{n+1}-a_n|<2^{-n}$ for all $n$ . How can I prove that $a_n$ is a cauchy sequence?

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marked as duplicate by C. Falcon, Elliot G, carmichael561, Community Dec 8 '16 at 20:14

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Hint: if $N\le m\le n$,

$$|a_m-a_n|\le |a_m-a_{m+1}|+|a_{m+1}-a_{m+2}|+\cdots+|a_{n-1}-a_n|< \sum_{k=m+1}^n \frac{1}{2^k}.$$

This sum is the tail of a convergent sum, and can be made arbitrarily small.

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By triangle inequality, we have that

$$\forall n,p > 0$$

$$|u_{n+p}-u_n|\leq \sum_{k=n}^{n+p-1}|u_{k+1}-u_k|$$ $$\leq \sum_{k=n}^{n+p-1}\frac{1}{2^k}$$

$$= \frac{1}{2^n}\frac{1-2^{-p}}{1-\frac{1}{2}}$$

$$\leq\frac{1}{2^{n-1}}.$$

and since $\lim_{n\to +\infty}\frac{1}{2^{n-1}}=0$,

we can conclude that $(u_n)_n$ is a Cauchy sequence.

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$|a_{n+1}−a_n|<2^{−n}$

$|a_{n+2}−a_n|\le|a_{n+2}-a_{n+1}|+ |a_n+1 - an| < 2^{-n}+2^{-n-1}$

assume $m>n$

$|a_m−a_n|\le \sum_\limits {i=n}^m 2^{-i}< 2^{-n+1}$

Let $N = (1-\log_2 \epsilon)$

For any $\epsilon, n,m> N \implies |a_m-a_n|<\epsilon$

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