# How to find sum of $n^{3}/n!$, when $n$ goes from 0 to infinity?

I know that i should get sums of $1/n!$ which are equal to $e$, but i don't know how to begin and how should i transform $n^{3}$?

See that \begin{align}{n^3\over n!} &= {n^2\over (n-1)!}\\ &= {n(n-1) \over (n-1)!} + {n \over (n-1)!}\\ &= {n\over (n-2)!} +{n\over (n-1)!}\\ &= {n-2\over (n-2)!}+{2\over (n-2)!} +{n-1\over (n-1)!} + {1\over (n-1)!}\\ &= {1\over(n-3)!} + {2\over(n-2)!} + {1\over(n-2)!} +{1\over(n-1)!}\\ &= {1\over(n-3)!} + {3\over(n-2)!} +{1\over(n-1)!} \end{align}

And use what you know for $\sum {1\over n!}$

Thus \begin{align} \sum_{n=0}^{+\infty} {n^3 \over n!} &= 1 + 4 +\sum_{n=3}^{+\infty} {1\over(n-3)!} + {3\over(n-2)!} +{1\over(n-1)!}\\ &= 5 +\sum_{n=0}^{+\infty} {1\over n!} + 3\sum_{n=1}^{+\infty} {1\over n!} +\sum_{n=2}^{+\infty} {1\over n!}\\ &= \sum_{n=0}^{+\infty} {1\over n!} + 3\left(1+\sum_{n=1}^{+\infty} {1\over n!} \right)+2+\sum_{n=2}^{+\infty} {1\over n!}\\ &= \sum_{n=0}^{+\infty} {1\over n!} + 3\sum_{n=0}^{+\infty} {1\over n!} +\sum_{n=0}^{+\infty} {1\over n!}\\ &= 5e \end{align}

The trick is to decompose $n^3$ as a sum of falling factorials,

$$n^3=n(n-1)(n-2)+an(n-1)+bn$$ that will simplify with the factors of $n!$

Identifying for two values of $n$ and solving

$$\begin{cases} 2^3=2(2-1)(2-2)+2(2-1)a+2b\\ 3^3=3(3-1)(3-2)+3(3-1)a+3b\\ \end{cases}$$

you obtain

$$n^3=n(n-1)(n-2)+3n(n-1)+n$$ and

$$\frac{n^3}{n!}=\frac1{(n-3)!}+\frac3{(n-2)!}+\frac1{(n-1)!}.$$

Actually the decomposition can be obtained easily as the system is triangular:

\begin{align}n^3-&\color{green}{n(n-1)(n-2)}=3n^2-2n,\\ 3n^2-2n-3&\color{green}{n(n-1)}=n,\\ n-&\color{green}{n}=0.\end{align}

This readily generalizes to any polynomial in $n$.

Hint: Expand $f(x)=e^x$ in a series and consider its derivatives.

• Hem: what new will the derivatives of $e^x$ bring ?
– user65203
Nov 5, 2016 at 16:59

You can consider the Euler operator $d_E=z\frac{d}{dz}$. Then $$\sum_{n\ge 0}\frac{n^3z^n}{n!}=(d_E)^3(e^z)$$ but, in general $$(z\frac{d}{dz})^n=\sum_{k=0}^n S_2(n,k)z^k(\frac{d}{dz})^k$$ where $S_2(n,k)$ are the Stirling numbers of the second kind. Here $$(z\frac{d}{dz})^3=z^3(\frac{d}{dz})^3+3\,z^2(\frac{d}{dz})^2+z(\frac{d}{dz})\ .$$ Applied to $e^z$, one gets $$\sum_{n\ge 0}\frac{n^3z^n}{n!}=(d_E)^3(e^z)=(z^3+3z^2+z)e^z$$ and specialization $z=1$ gives $\sum_{n\ge 0}\frac{n^3}{n!}=5e$.

This method gives, in general,
$$\sum_{n\ge 0}\frac{n^m}{n!}=B_m\,e$$ where $B_m$ is the $m$-th Bell number (Dobiński's formula).

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

In general, with $\ds{k \in \mathbb{N}_{\ \geq\ 0}}$:

\begin{align} \sum_{n = 0}^{\infty}{n^{k} \over n!} & = \sum_{n = 0}^{\infty}{1 \over n!}\,\ \overbrace{k!\oint_{\verts{z}\ =\ 1^{-}}{\expo{nz} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}}^{\ds{n^{k}}}\ =\ k!\oint_{\verts{z}\ =\ 1^{-}}{1 \over z^{k + 1}} \sum_{n = 0}^{\infty}{\pars{\expo{z}}^{n} \over n!}\,{\dd z \over 2\pi\ic} \\[5mm] & = k!\oint_{\verts{z}\ =\ 1^{-}}{\exp\pars{\expo{z}} \over z^{k + 1}} \,{\dd z \over 2\pi\ic} = \bbox[#ffe,10px,border:1px dotted navy]{\ds{k!\bracks{z^{k}}\exp\pars{\expo{z}}}} \end{align} The $\ds{\exp\pars{\expo{z}}}$-Taylor expansion is rather cumbersome for high values of $\ds{k}$ albeit it can be evaluated with some CAS. Namely, \begin{align} \sum_{n = 0}^{\infty}{n^{k} \over n!} & = k!\bracks{z^{k}}\bracks{\expo{}\color{#f00}{z^{0}} + \expo{}\color{#f00}{z^{1}} + \expo{}\color{#f00}{z^{2}} + {{\atop{\atop{\atop{\atop{\ds{5\expo{} \over 6}}}}}} \atop \uparrow}\!\!\!\color{#f00}{z^{3}} + {5\expo{} \over 8}\,\color{#f00}{z^{4}} + {13\expo{} \over 30}\,\color{#f00}{z^{5}} + {203\expo{} \over 720}\,\color{#f00}{z^{6}} + \,\mrm{O}\pars{\color{#f00}{z^{7}}}} \end{align}

$$\sum_{n = 0}^{\infty}{n^{3} \over n!} = 3!\,{5\expo{} \over 6} = \ \bbox[#ffe,10px,border:1px dotted navy]{\ds{{5\expo{}}}}$$