limit of $n \left(e-\sum_{k=0}^{n-1} \frac{1}{k!}\right) = ?$ 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.

I will show that
$\lim_{n \to \infty} n^m(e-\sum_{k=0}^{n-1} \frac{1}{k!})
= 0
$
for any finite $m$.
$\begin{array}\\
e-\sum_{k=0}^{n-1} \frac{1}{k!}
&=\sum_{k=n}^{\infty} \frac{1}{k!}\\
&=\frac1{n!}\sum_{k=n}^{\infty} \frac{n!}{k!}\\
&=\frac1{n!}\sum_{k=n}^{\infty} \frac{1}{\prod_{j=n+1}^{k} j}\\
&=\frac1{n!}\sum_{k=n}^{\infty} \frac{1}{\prod_{j=1}^{k-n} (j+n)}\\
&\le\frac1{n!}\sum_{k=n}^{\infty} \frac{1}{\prod_{j=1}^{k-n} (1+n)}\\
&=\frac1{n!}\sum_{k=n}^{\infty} \frac{1}{(n+1)^{k-n}}\\
&=\frac1{n!}\sum_{k=0}^{\infty} \frac{1}{(n+1)^{k}}\\
&=\frac1{n!}\frac{1}{1-\frac1{n+1}}\\
&=\frac1{n!}\frac{n+1}{n}\\
\end{array}
$
Since
$\lim_{n \to \infty} \frac{n^m}{n!}
= 0
$
for any finite $m$,
$\lim_{n \to \infty} n^m(e-\sum_{k=0}^{n-1} \frac{1}{k!})
= 0
$
for any finite $m$.
Also note that
$\lim_{n \to \infty} n!(e-\sum_{k=0}^{n-1} \frac{1}{k!})
= 1
$.
