已知

\sum_{n = 0}^{\infty} \frac {1}{n!} = e \\

利用泰勒展开很容易证明，并且也很容易得到

\begin{aligned} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k !}&=\frac{1}{e}\\ \sum_{k=1}^{\infty} \frac{k}{(2 k+1) !}&=\frac{1}{2 e}\\ \sum_{k=0}^{\infty} \frac{1}{(2 k) !}&=\frac{1}{2}\left(e+\frac{1}{e}\right)\\ \sum_{k=0}^{\infty} \frac{1}{(2 k+1) !}&=\frac{1}{2}\left(e-\frac{1}{e}\right) \end{aligned}\\

接下来主要讨论以下几个拓展

\sum_{n=0}^{\infty}\frac{1}{(n+a)!}的计算

\begin{aligned} &\sum_{n=0}^{\infty}\frac{1}{(n+a)!}=\lim_{m\to\infty}\sum_{n=0}^{m}\frac{1}{(n+a)!}\\ &=\lim_{m\to\infty}\frac{e\left(a!\Gamma(a+m+1,1)-a\Gamma(a,1)(a+m)!\right)}{a!(a+m)!}\\ &=\frac{e}{a!}\lim_{m\to\infty}\frac{a!\Gamma(a+m+1,1)-a\Gamma(a,1)(a+m)!}{(a+m)!}\\ &=\frac{e}{a!}\lim_{m\to\infty}\left(\frac{a!\Gamma(1+a+m,1)}{(a+m)!}-a\Gamma(a,1)\right)\\ &=\frac{e}{a!}\left(\lim_{m\to\infty}\frac{a!\Gamma(1+a+m,1)}{(a+m)!}-a\Gamma(a,1)\right)\\ &=\frac{e}{a!}\left(a!\lim_{m\to\infty}\frac{\Gamma(1+a+m,1)}{(a+m)!}-a\Gamma(a,1)\right)\\ &=\frac{e}{a!}\left(a!\cdot1-a\Gamma(a,1)\right)\\ &=e-\frac{e\Gamma(a,1)}{\Gamma(a)}\\ \end{aligned}\\ \\\begin{aligned} \sum \limits_{n=0}^{\infty} \frac{1}{(n+1)!}&= \sum \limits_{n=1}^{\infty} \frac{1}{n!} =\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+ \dots=e-1\\ \sum \limits_{n=0}^{\infty} \frac{1}{(n+2)!}&= \sum \limits_{n=2}^{\infty} \frac{1}{n!} =\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots=e-2\\ \sum \limits_{n=0}^{\infty} \frac{1}{(n+3)!}&= \sum \limits_{n=3}^{\infty} \frac{1}{n!} =\frac{1}{3!}+\frac{1}{4!}+\dots=e-\frac{5}{2} \end{aligned} \\

\sum \limits_{n=0}^{\infty} \frac{1}{(kn)!}的计算

如果\frac nk\not\in\mathbb{Z}拓展感想，则

\sum_{j=0}^{k-1}e^{2\pi ij\frac nk}=\frac{e^{2\pi in}-1}{e^{2\pi i\frac nk}-1}=0 \\

如果\frac nk\in\mathbb{Z}拓展感想，则

\sum_{j=0}^{k-1}e^{2\pi ij\frac nk}=\sum_{j=0}^{k-1}1=k \\

所以

\begin{aligned} \sum_{n=0}^\infty\frac1{(kn)!} &=\sum_{n=0}^\infty\overbrace{\left(\frac1k\sum_{j=0}^{k-1}e^{2\pi ij\frac nk}\right)}^{1\iff\frac nk\in\mathbb{Z}}\frac1{n!}\\ &=\frac1k\sum_{j=0}^{k-1}\sum_{n=0}^\infty\frac{\left(e^{2\pi ij/k}\right)^n}{n!}\\ &=\frac1k\sum_{j=0}^{k-1}e^{\large e^{2\pi ij/k}}\\ &=\frac1k\sum_{j=0}^{k-1}e^{\cos(2\pi j/k)+i\sin(2\pi j/k)}\\ &=\frac1k\sum_{j=0}^{k-1}e^{\cos(2\pi j/k)}\left(\cos(\sin(2\pi j/k))+i\sin(\sin(2\pi j/k))\right)\\ &=\frac1k\sum_{j=0}^{k-1}e^{\cos(2\pi j/k)}\cos(\sin(2\pi j/k)) \end{aligned}\\

补充：

\begin{aligned} \sum_{n=1}^\infty{n^2\over n!}&=\sum_{n=1}^\infty{n\over(n-1)!}=\sum_{m=0}^\infty{m+1\over m!}=\sum_{m=0}^\infty{m\over m!}+e\\ &=\sum_{m=1}^\infty{m\over m!}+e=\sum_{m=1}^\infty{1\over(m-1)!}+e\\ &=\sum_{k=0}^\infty{1\over k!}+e=e+e\\ &=2e \end{aligned} \\

\begin{align} \sum_1^{\infty} \frac{n^3}{n!}&=\sum_1^{\infty} \frac{n (n - 1) (n - 2) + 3 n (n - 1) + n}{n!}\\\\ &=\sum_3^{\infty} \frac{n (n - 1) (n - 2)}{n!}+3\sum_2^{\infty} \frac{n(n - 1) }{n!}+\sum_1^{\infty} \frac{n}{n!}\\\\ &=\sum_3^{\infty} \frac{1}{(n-3)!}+3\sum_2^{\infty} \frac{1}{(n-2)!}+\sum_1^{\infty} \frac{1}{(n-1)!}\\\\ &=\sum_0^{\infty} \frac{1}{n!}+3\sum_0^{\infty} \frac{1}{n!}+\sum_0^{\infty} \frac{1}{n!}\\\\ &=5e \end{align}\\

思考：

\Large\color{red}{\sum\limits_{n=1}^\infty \frac{n^k}{n!}} \quad k\text{为大于3的正整数}\\

彩蛋：

\Large \sum_{n=0}^{\infty}\frac{\left(n!\right)^2}{\left(2n\right)!}=\frac{2}{27}\left(18+\sqrt{3}\pi\right)\\

参考：

[01]Formula for $\sum \limits_{n=0}^{\infty} \frac{1}{(n+a)!}$

[02]/questions/1708900/sum-of-sum-limits-n-0-infty-frac1kn?rq=1

[03]/questions/548563/calculate-sum-limits-k-0-infty-frac12k-choose-k

[04]Table of Integrals, Series, and Products (Eighth Edition)[M]. 2014.