<s>[INST] Show that the sequence $(a_n)$ where $a_n = 1 + \frac{1}{2!} + \ldots + \frac{1}{n!}$ is of Cauchy. 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 $m$. 
 [/INST] 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 
$$m>n>n_0\quad\implies\quad|a_m-a_n|<\varepsilon$$
</s>