Consider a sequence $(x_n)$ such that $\forall n \in \mathbb N, |x_{n+1}-x_{n}|\leq 2^{-n}$

How to show that it converges? Any hints would be appreciated


I assume that you are speaking abot real analysis.

For any $\epsilon>0$, there exists $n_0\in\Bbb N$ such that $1/2^{n_0-1}<\epsilon$. For any natural numbers $q>p\geq n_0$ you have $$|x_q-x_p|\leq\sum_{k=p}^{q-1}|x_{k+1}-x_k|\leq \sum_{k=p}^{q-1}\frac 1{2^k}\leq\frac1{2^{p-1}}<\epsilon$$

Therefore, the sequence is a Cauchy sequence and hence converges.

  • $\begingroup$ yeah I thought about Cauchy but didn't realise how to use it. $\endgroup$ – user2965303 Jun 10 '14 at 13:47

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