Find the value of $\sum_{n=0}^{\infty} \left(e-\sum_{k=0}^{n} \frac{1}{k!}\right).$ We know $$e=\sum_{k=0}^{\infty} \frac 1{k!}.$$ But in the question the upper bound of the second sum is $n$ so we can't write $e$ directly. Since upper bound of $n$ is $\infty$ so tried to split the summations to get a clue. The given sum is:
$\left(e-\frac 1{0! }\right) +\left[e-\left(\frac 1{0!}+\frac{1}{1! }\right) \right]+\cdots+\left[e-\underbrace{\left(\frac{1}{0! }+\frac{1}{1! }+\frac 1{2! }+\cdots\infty\right)}_{=e}\right].$
Now I'm clueless. The last few terms tend to $0$ but how to handle the other terms? Please suggest something. I haven't dealt with infinite sums that much, I have just started to deal with such expressions in this lockdown period. So there may be something which hasn't striked me yet.. Please help! And yeah, the answer is $e$.
 A: Your sum equals
$$
\sum\limits_{n = 0}^\infty  {\sum\limits_{k = n + 1}^\infty  {\frac{1}{{k!}}} }  = \sum\limits_{n = 1}^\infty  {n \cdot \frac{1}{{n!}}}  = \sum\limits_{n = 1}^\infty  {\frac{1}{{(n - 1)!}}}  = \sum\limits_{n = 0}^\infty  {\frac{1}{{n!}}}  = e.
$$
A: Change order of summation
$$
\sum^\infty_{n=0}\sum^\infty_{m>n}\frac{1}{m!}=\sum^\infty_{m=1}\sum^{m-1}_{n=0}\frac{1}{m!}=\sum^\infty_{m=1}\frac{m}{m!}=\sum^\infty_{m=0}\frac{1}{m!}=e$$
A: Let $ n $ be a positivie integer, note that : $\mathrm{e}-\sum\limits_{k=0}^{n}{\frac{1}{k!}}=\frac{\mathrm{e}}{n!}\int_{0}^{1}{x^{n}\mathrm{e}^{-x}\,\mathrm{d}x}\cdot $
Since : \begin{aligned}\sum_{n=0}^{p}{\frac{1}{n!}\int_{0}^{1}{x^{n}\mathrm{e}^{-x}\,\mathrm{d}x}}-\int_{0}^{1}{\sum_{n=0}^{+\infty}{\frac{x^{n}}{n!}\mathrm{e}^{-x}}\,\mathrm{d}x}=\int_{0}^{1}{\sum_{n=p+1}^{+\infty}{\frac{x^{n}}{n!}\mathrm{e}^{-x}}\,\mathrm{d}x}\underset{p\to +\infty}{\longrightarrow}0\end{aligned}
We get that : \begin{aligned}\sum_{n=0}^{+\infty}{\frac{1}{n!}\int_{0}^{1}{x^{n}\mathrm{e}^{-x}}\,\mathrm{d}x}=\lim_{p\to +\infty}{\sum_{n=0}^{p}{\frac{1}{n!}\int_{0}^{1}{x^{n}\mathrm{e}^{-x}\,\mathrm{d}x}}}=\int_{0}^{1}{\sum_{n=0}^{+\infty}{\frac{x^{n}}{n!}\mathrm{e}^{-x}}\,\mathrm{d}x}=1\end{aligned}
Thus : \begin{aligned}\sum_{n=0}^{+\infty}{\left(\mathrm{e}-\sum_{k=0}^{n}{\frac{1}{k!}}\right)}=\mathrm{e}\sum_{n=0}^{+\infty}{\frac{1}{n!}\int_{0}^{1}{x^{n}\mathrm{e}^{-x}}\,\mathrm{d}x}=\mathrm{e}\end{aligned}
