I am reading a proof in which the following result is assumed.
Let $(x_n)_{n\in\mathbb{N}}$ be a real valued sequence such that $$ \limsup_{m\rightarrow\infty}\frac{1}{m}\log|x_{m+1}-x_m| < 0. $$ Then $(x_n)_{n\in\mathbb{N}}$ is a Cauchy sequence.
Could someone help me with proving this. Thank you in advance.
If $\limsup_{m\rightarrow\infty}\frac{1}{m}\log\lvert x_{m+1}-x_m| < 0$, then $$ \frac{1}{m}\log\lvert x_{m+1}-x_m| < -a, \quad\text{whenever $m\ge n_0$}, $$ for some $a>0$ and for some $n_0\in\mathbb N$. Thus $$ \lvert x_{m+1}-x_m|<\mathrm{e}^{-ma}, \quad\text{whenever $m\ge n_0$}, $$ and hence $$ \lvert x_m-x_n|\le \lvert x_m-x_{m-1}|+\cdots+\lvert x_{n+1}-x_n|\le \sum_{k=n}^{m-1}\mathrm{e}^{-ka}\le \sum_{k=n}^{\infty}\mathrm{e}^{-ka}=\frac{\mathrm{e}^{-na}}{1-\mathrm{e}^{-a}} $$ whenever $m\ge n\ge n_0$.