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

The binomial coefficient

$${n+m\choose n}={(n+m)!\over n!m!}$$

is a positive integer if $m,n\ge0$, which implies

$${1\over(n+m)!}\le{1\over n!m!}$$

$$e-\sum_{k=0}^{n-1}{1\over k!}=\sum_{k=n}^\infty{1\over k!}=\sum_{m=0}^\infty{1\over(n+m)!}\le{1\over n!}\sum_{m=0}^\infty{1\over m!}={e\over n!}$$

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 $0$.

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

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

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