2022-01-17

Whats

Raymond Foley

Expert

The value of ${T}_{n}\phantom{\rule{0.222em}{0ex}}=\sum _{k=1}^{\mathrm{\infty }}\frac{{k}^{n}}{k!}$ is ${B}_{n}\cdot e$, where ${B}_{n}$ is the n-th Bell number.
To see this, note that
${T}_{n+1}=\sum _{k=1}^{\mathrm{\infty }}\frac{{k}^{n+1}}{k!}=\sum _{k=0}^{\mathrm{\infty }}\frac{{\left(k+1\right)}^{n}}{k!}$
$=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}\sum _{j=0}^{n}\left(\begin{array}{c}n\\ j\end{array}\right){k}^{j}$
$=\sum _{j=0}^{n}\left(\begin{array}{c}n\\ j\end{array}\right)\sum _{k=1}^{\mathrm{\infty }}\frac{{k}^{j}}{k!}$
$=\sum _{j=0}^{n}\left(\begin{array}{c}n\\ j\end{array}\right){T}_{j}$
This is precisely the recursion formula that the Bell numbers follow, except that every term here is being multiplied by e.

Beverly Smith

Expert

A basic technique in real (complex) analysis is term by term differentiation of power series:
${e}^{z}=\sum _{k=0}^{\mathrm{\infty }}\frac{{z}^{k}}{k!}$
${e}^{z}={\left({e}^{z}\right)}^{\prime }=\sum _{k=1}^{\mathrm{\infty }}k\cdot \frac{{z}^{k-1}}{k!}$
${e}^{z}=\left({e}^{z}\right){}^{″}=\sum _{k=1}^{\mathrm{\infty }}k\left(k-1\right)\frac{{z}^{k-2}}{k!}$
Evaluating at $z=1$, one immediately has
$e=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}$
$e=\sum _{k=1}^{\mathrm{\infty }}k\cdot \frac{1}{k!}$
$e=\sum _{k=1}\left({k}^{2}-k\right)\frac{1}{k!}$
Combining the second and third equalities, we have the answer.

alenahelenash

Expert

Just to give a slightly different approach, $\sum _{n=1}^{\mathrm{\infty }}\frac{{n}^{2}}{n!}=\sum _{n=1}^{\mathrm{\infty }}\frac{n}{\left(n-1\right)!}$ $\sum _{m=0}^{\mathrm{\infty }}\frac{m+1}{m!}=\sum _{m=0}^{\mathrm{\infty }}\frac{m}{m!}+e$ $\sum _{m=1}^{\mathrm{\infty }}\frac{m}{m!}+e=\sum _{m=1}^{\mathrm{\infty }}\frac{1}{\left(m-1\right)!}+e$ $=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}+e=e+e$