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.
 A: The binomial coefficient 
$${n+m\choose n}={(n+m)!\over n!m!}$$ 
is a positive integer if $m,n\ge0$, which implies 
$${1\over(n+m)!}\le{1\over n!m!}$$ 
so
$$e-\sum_{k=0}^{n-1}{1\over k!}=\sum_{k=n}^\infty{1\over k!}=\sum_{m=0}^\infty{1\over(n+m)!}\le{1\over n!}\sum_{m=0}^\infty{1\over m!}={e\over n!}$$
A: You can try to show that $\frac{1}{n!}\leq\frac{2}{2^n}$, then $\frac{1}{n!}+\frac{1}{(n+1)!}+\dots\leq\frac{4}{2^n}$.
$$\frac{1}{n!}\leq\frac{1}{1\times2\times3\times\dots\times n}\leq\frac{1}{1\times2\times2\times\dots\times2}=\frac{2}{2^n}$$
So $\frac{1}{n!}+\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\dots\leq\frac{2}{2^n}+\frac{2}{2^{n+1}}+\frac{2}{2^{n+2}}+\dots\leq\frac{4}{2^n}$.
Hence,
$$\lim_{n\rightarrow\infty}n(e-\sum_{k=0}^{n-1}\frac{1}{k!})\leq\lim_{n\rightarrow\infty}\frac{4n}{2^n}=0$$
Also, it is clear that the limit is non-negative, as $e-\sum_{k=0}^{n-1}\frac{1}{k!}\geq0$.
Hence, the limit of the expression as $n\rightarrow\infty$ is $0$.
A: 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
$$0\leq n(e-\sum_{k=0}^{n-1} \frac{1}{k!}) = n\frac{e^{\xi_n}}{n!} \leq \frac{e}{(n-1)!} \stackrel{n \to \infty}{\longrightarrow}0$$
A: HINT: (if you're familiar with the gamma function):
$$\lim_{n\to\infty}n\left(e-\sum_{k=0}^{n-1}\frac{1}{k!}\right)=$$
$$\lim_{n\to\infty}n\left(e-\frac{e\Gamma(n,1)}{\Gamma(n)}\right)=$$
$$\lim_{n\to\infty}n\left(\frac{e\left(\Gamma(n)-\Gamma(n,1)\right)}{\Gamma(n)}\right)=$$
$$\lim_{n\to\infty}\frac{en\left(\Gamma(n)-\Gamma(n,1)\right)}{\Gamma(n)}=$$
$$e\lim_{n\to\infty}\frac{n\Gamma(n)-n\Gamma(n,1)}{\Gamma(n)}=$$
$$e\lim_{n\to\infty}\left(n-\frac{n\Gamma(n,1)}{\Gamma(n)}\right)=$$
$$e\lim_{n\to\infty}\left(n-\frac{n^2\Gamma(n,1)}{n!}\right)=$$
$$-e\lim_{n\to\infty}\frac{n^2\Gamma(n,1)-nn!}{n!}$$
A: 
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.
A: 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
$.
