<s>[INST] Question on Rudin Theorem 3.31 Proof On page 65, we have:

$e-s_n = \frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}+\dots < \frac{1}{(n+1)!}(1 + \frac{1}{n+1}+\frac{1}{(n+1)^2}+\dots)=\frac{1}{n!n}$

Given $s_n = \sum_{k=0}^{n}{\frac{1}{k!}}$.
What I don't understand is, where does $\frac{1}{n!n}$ come from?
 [/INST] Note that since for $ |a| < 1$ (Geometric series formula)
$$
\sum_{i=0}^{\infty}a^i = \frac{1}{1-a},
$$
choosing $a := \frac{1}{n+1}$, on the right hand side it follows that
$$
\frac{1}{(n+1)!}\left(\sum_{i=0}^{\infty}\left(\frac{1}{n+1}\right)^i\right) = \frac{1}{(n+1)!}\left(\frac{1}{1 - \frac{1}{n+1}}\right) = \frac{1}{n!n}
$$
</s>