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.


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.