Proving that a sequence is convergent

Question

I have to prove that for a sequence $\{a_n\}$ with $|a_{n+1} - a_n| < 2^{-n}$ is convergent.

So I thought, that if a be a series then it will convergence against $\dfrac{1}{2^n}$ and for $n>1$ the series is convergent which means, that the sequence has to be convergent too.

hint: You prove such a sequence to be Cauchy sequence, and this implies convergence. — DeepSea, Jan 16 '17 at 06:24
See also: Proving that a sequence such that $|a_{n+1} - a_n| \le 2^{-n}$ is Cauchy and other posts linked there. — Martin Sleziak, Jan 16 '17 at 07:43

score 3 · Accepted Answer · answered Jan 16 '17 at 06:31

Fix $\epsilon > 0$. Now choose $N$ so large that $\sum_{k=N}^{\infty} \frac{1}{2^k} < \epsilon$. For any $m,n > N$, we have \begin{align*} |a_m - a_n| \leq \sum_{k=n+1}^m |a_k - a_{k-1}| < \sum_{k=n+1}^m \frac{1}{2^k} < \sum_{k=N}^{\infty} \frac{1}{2^k} < \epsilon \end{align*} So $\{a_n\}$ is Cauchy. Assuming the $a_i$'s belong to a complete metric space, the sequence must converge.

Proving that a sequence is convergent

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