TITLE: limit of $n \left(e-\sum_{k=0}^{n-1} \frac{1}{k!}\right) = ?$
QUESTION [6 upvotes]: As in the title:
$$
\lim_{n\to\infty} n \left(e-\sum_{k=0}^{n-1} \frac{1}{k!}\right) = ?
$$
Numerically, it seems 0, but how to prove/disprove it?
I tried to show that the speed of convergence of the sum to e is faster than $1/n$ but with no success.

REPLY [2 votes]: Late answer because something similar was tagged as "duplicate" (although it is not that what a "duplicate" is).
So,I adjust the answer from there to the question here:
Taylor gives

$e^1 = \sum_{k=0}^{n-1}\frac{1}{k!} + \frac{e^{\xi_n}}{n!}$ with $\xi_n \in (0,1)$

It follows
$$0\leq n(e-\sum_{k=0}^{n-1} \frac{1}{k!}) = n\frac{e^{\xi_n}}{n!} \leq \frac{e}{(n-1)!} \stackrel{n \to \infty}{\longrightarrow}0$$