Show that the sequence $(a_n)$ where $$a_n = 1 + \frac{1}{2!} + \ldots + \frac{1}{n!}$$ is of Cauchy.
Comments: I will use the definition $$\forall \epsilon > 0, \exists n_0 \in \mathbb{N}; n,m > n_0 \Rightarrow |a_n - a_m|< \epsilon.$$ If $n>m$ $$|a_n - a_m| = \left|\frac{1}{(n+1)!} + \ldots + \frac{1}{n!}\right| \leq \left|\frac{1}{(n+1)!}\right| + \ldots + \left|\frac{1}{n!}\right|$$ I do not know how to limit the $\epsilon$ and define $n_0$ without it depends of $n$ and $m$.
If $m>n>1$, then \begin{align} |a_m-a_n|&=\left|\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\ldots+\frac{1}{m!}\right|\\&<\left|\frac{1}{(n+1)!}+\frac{1}{2}\frac{1}{(n+1)!}+\ldots+\frac{1}{2^{m-n-1}}\frac{1}{(n+1)!}\right|\\ &=\frac{1}{(n+1)!}\sum_{k=0}^{m-n-1}\frac{1}{2^k}\\ &<\frac{1}{(n+1)!}\sum_{k=0}^{\infty}\frac{1}{2^k}\\ &=\frac{2}{(n+1)!}\\ &<\frac{1}{n!} \end{align} Then, given $\varepsilon>0$, by taking $n_0$ such that $n_0!>\max\left(\frac{1}{\varepsilon},1\right)$, we have $$m>n>n_0\quad\implies\quad|a_m-a_n|<\varepsilon$$