I am given a sequence, $s_{n}$, and told that, $\forall n\in\mathbb{N}$, $|s_{n+1}-s_{n}|<\frac{1}{2^{n}}$. I am asked to prove that $s_n$ is Cauchy. Is the following headed in the right direction?
Suppose WLOG that $m=n+k, k\in \mathbb{N}$. So $|s_{m+k}-s_{n}|\leq|s_{n+k}-s_{n+k-1}|+\cdot \cdot \cdot |s_{n+1}-s_{n}|$ (by the triangle inequality) $< \frac{1}{2^{n+k-1}}+\cdot \cdot \cdot +\frac{1}{2^{n}}<\frac{k}{2^{n}}$ (k lots of $\frac{1}{2^{n}}$, the largest term in the sum on the left hand side) < $\frac{k}{n}$. So, choose $N=\frac{k}{\epsilon}$, so that $m>n>N \implies|s_{m+k}-s_{n}|<\epsilon.$
Thank you.
Your idea is good, but your $N$ isn't allowed to depend on $k$, because then our $N$ will be depend on $m,n$ and that is bad.
Just think of the geometric series. For $m>n\geq N$ we have $$ |s_m-s_n|\leq \sum_{k=n}^{m-1} |s_{k+1}-s_k|\leq \sum_{k=n}^{m-1} \frac{1}{2^k}\leq \sum_{k=n}^{\infty}\frac{1}{2^k}=\frac{1}{2^n}\sum_{k=0}^{\infty}\frac{1}{2^k}=\frac{1}{2^{n-1}}\leq \frac{1}{2^{N-1}}. $$