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 tried to show that the speed of convergence of the sum to e is faster than $1/n,$ but with no success.

Have you tried Stirling's approximation ?

In my opinion, a far more interesting question would have been trying to prove that 

$$\lim_{n\to\infty}(n+a)\bigg[~e^b-\bigg(1+\dfrac b{n+c}\bigg)^{n+d}~\bigg]~=~b~e^b~\bigg(\dfrac b2+c-d\bigg),$$

since, in this case, the growth of the latter formula towards $e^b$ is comparable to that of $1/n.$  In your example, however, in order for the product to converge to a “meaningful” non-zero  quantity, the order of the multiplication factor should have been somewhere in the range of  $n!$, as has already been pointed put by Daniel Fischer in the comments.
