Let $s_{n}$ be a sequence such that $|s_{n+1}-s_{n}|<2^{-n}$. Prove that $s_{n}$ is Cauchy. Is this proof sound?
Proof: Let $\epsilon >0$. Since the sequence $\frac{1}{2^{n}}$ converges to $0$, there exists $N$ such that $n>N$ implies $\left|\frac{1}{2^{n}}-0\right|=\frac{1}{2^{n}}<\frac{\epsilon}{2}$. Assume without loss of generality $n,m>N$ with $n\geq m$. Then \begin{align*}
|s_{n}-s_{m}|&\leq |s_{n}-s_{n-1}|+|s_{n-1}-s_{n-2}|+...+|s_{m+2}-s_{m-1}|+|s_{m-1}+s_{m}|\\
&<\frac{1}{2^{n-1}}+\frac{1}{2^{n-2}}+...+\frac{1}{2^{m-1}}+\frac{1}{2^{m}}\\
&=\frac{\frac{1}{2^{m}}\left(1-\left(\frac{1}{2}\right)^{n-m}\right)}{\frac{1}{2}}\text{$\quad$ by the geometric series formula}\\
&=\frac{\frac{1}{2^{m}}-\frac{1}{2^{n}}}{\frac{1}{2}}\\
&=\frac{2}{2^{m}}-\frac{2}{2^{n}}\\
&<\frac{2}{2^{m}}\\
&=2\left(\frac{1}{2^{m}}\right)\\
&<2\left(\frac{\epsilon}{2}\right)\\
&=\epsilon.
\end{align*}

It's right, and the condition can be more looser. We just need $ |s_{n+1}-s_{n}|\leqslant n^{-2} $. The key steps of proof are
\begin{align*}
    |s_{n+p}-s_{n}| & \leqslant |s_{n+p}-s_{n+p-1}|+\dots+|s_{n+1}-s_{n}|\\
    & \leqslant\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}+\dots+\frac{1}{(n+p-1)^{2}}\\
    & \leqslant\frac{1}{(n-1)n}+\frac{1}{n(n+1)}+\dots+\frac{1}{(n+p-2)(n+p-1)}\\
    & \leqslant \frac{1}{n-1}-\frac{1}{n}+\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n+p-2}+\frac{1}{n+p-1}\\
    & \leqslant \frac{1}{n-1}-\frac{1}{n+p-1}<\frac{1}{n-1}\to 0,\qquad n\to\infty
\end{align*}
