Sum of the series $\sum _{ r=1 }^{ \infty }{ \frac { { r }^{ 3 } }{ r! } } $

Question

How do I prove that the sum of the series $1+\frac { { 2 }^{ 3 } }{ 2! } +\frac { { 3 }^{ 3 } }{ 3! } +\frac { { 4 }^{ 3 } }{ 4! } +.... = 5e$

Answer 1

Note that $r^3=r((r-1)(r-2)+3(r-1)+1)$. Hence $$\sum _{ r=1 }^{ \infty }\frac {r^{ 3 } }{ r! }= \sum _{ r=1 }^{ \infty }\frac {r(r-1)(r-2) }{ r! } +3\sum _{ r=1 }^{ \infty }\frac {r(r-1)}{ r! }+\sum _{ r=1 }^{ \infty }\frac {r}{ r! }.$$ Can you take it from here? Show your efforts!

Answer 2

hint

$$\sum_{n=1}^\infty \frac {n^3}{n!}=$$

$$\sum_{n=1}\frac {n^2-1+1}{(n-1)!}=$$ $$\sum_{n=2} \frac {n-2+3}{(n-2)!}+e=$$

$$\sum_{n=3} \frac {1}{(n-3)!}+3e+e=$$ $$5e $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{r = 1}^{\infty}{r^{3} \over r!} & = \left.\pars{x\,\totald{}{x}}^{3}\sum_{r = 1}^{\infty}{x^{r} \over r!} \right\vert_{\ x\ =\ 1} = \left.\pars{x\,\totald{}{x}}^{3}\pars{\expo{x} - 1} \right\vert_{\ x\ =\ 1} = \bbx{5\expo{}} \end{align}