Maximus George

2022-01-21

Evaluate $\sum _{k=1}^{\mathrm{\infty }}\frac{{k}^{2}}{\left(k-1\right)!}$

sphwngzt

Let's start ${e}^{x}=\sum _{k=1}^{\mathrm{\infty }}\frac{{x}^{k-1}}{\left(k-1\right)!}$ then:
$x{e}^{x}=\sum _{k=1}^{\mathrm{\infty }}\frac{{x}^{k}}{\left(k-1\right)!}$
$x{\left(x{e}^{x}\right)}^{\prime }=x{e}^{x}+{x}^{2}{e}^{x}=\sum _{k=1}^{\mathrm{\infty }}\frac{k{x}^{k}}{\left(k-1\right)!}$
$x{\left(x{\left(x{e}^{x}\right)}^{\prime }\right)}^{\prime }=x\left({e}^{x}+3x{e}^{x}+{x}^{2}{e}^{x}\right)=\sum _{k=1}^{\mathrm{\infty }}\frac{{k}^{2}{x}^{k}}{\left(k-1\right)!}$
Set $x=1$ to get:
$5e=\sum _{k=1}^{\mathrm{\infty }}\frac{{k}^{2}}{\left(k-1\right)!}$

jojoann325

The most elementary calculation is probably this one:
$\sum _{k\ge 0}\frac{{\left(k+1\right)}^{2}}{k!}=1+\sum _{k\ge 1}\frac{{k}^{2}+2k+1}{k!}$
$=1+\sum _{k\ge 1}\frac{k}{\left(k-1\right)!}+2\sum _{k\ge 1}\frac{1}{\left(k-1\right)!}+\sum _{k\ge 1}\frac{1}{k!}$
$=\sum _{k\ge 1}\frac{k-1+1}{\left(k-1\right)!}+2\sum _{k\ge 0}\frac{1}{k!}+\sum _{k\ge 0}\frac{1}{k!}$
$=1+\sum _{k\ge 2}\frac{1}{\left(k-2\right)!}+\sum _{k\ge 2}\frac{1}{\left(k-1\right)!}+3\sum _{k\ge 0}\frac{1}{k!}$
$=\sum _{k\ge 0}\frac{1}{k!}+4\sum _{k\ge 0}\frac{1}{k!}$
$=5\sum _{k\ge 0}\frac{1}{k!}$
$=5e$.
One can also make use of the identity
${x}^{n}=\sum _{k}\left(\begin{array}{c}3\\ i\end{array}\right){k}^{i}$
$=0\cdot {k}^{0}+1\cdot {k}^{1}+3\cdot {k}^{2}+1\cdot {k}^{3}$
$=k+3k\left(k-1\right)+k\left(k-1\right)\left(k-2\right)$
Whence
$\frac{{k}^{2}}{\left(k-1\right)!}=\frac{{k}^{3}}{k!}=\frac{k+3k\left(k-1\right)+k\left(k-1\right)\left(k-2\right)}{k!}$
$=\frac{1}{\left(k-1\right)!}+\frac{3}{\left(k-2\right)!}+\frac{1}{\left(k-3\right)!}$,
and summing over l yields 5e as before.

RizerMix

For $\frac{P\left(n\right)}{\left(n-r\right)!}$ where $P\left(n\right)$ is a polynomial. If the degree of $P\left(n\right)$ is $m>0$, we can write $P\left(n\right)={A}_{0}+{A}_{1}\left(n-r\right)+{A}_{2}\left(n-r\right)\left(n-r-1\right)+...+{A}_{m}\left(n-r\right)\left(n-r-1\right)...\left\{\left(n-r\right)-\left(m-1\right)\right\}$ Here ${k}^{2}=C+B\left(k-1\right)+A\left(k-1\right)\left(k-2\right)$ Putting $k=1$ in the above identity, $C=1$ $k=0⇒2A-B+C=0,2A=B-C=3-1⇒A=1$ or comparing the coefficients of So, $\frac{{k}^{2}}{\left(k-1\right)!}=\frac{\left(k-1\right)\left(k-2\right)+3\left(k-1\right)+1}{\left(k-1\right)!}=\frac{1}{\left(k-3\right)!}+\frac{3}{\left(k-2\right)!}+\frac{1}{\left(k-1\right)!}$ $\sum _{k=1}^{\mathrm{\infty }}\frac{{k}^{2}}{\left(k-1\right)!}=\sum _{k=1}^{\mathrm{\infty }}\left(\frac{1}{\left(k-3\right)!}+\frac{3}{\left(k-2\right)!}+\frac{1}{\left(k-1\right)!}$ $\sum _{k=1}^{\mathrm{\infty }}\frac{{k}^{2}}{\left(k-1\right)!}=\sum _{k=1}^{\mathrm{\infty }}\left(\frac{\left(k-1\right)\left(k-2\right)+3\left(k-1\right)+1}{\left(k-1\right)!}\right)$ $=\sum _{k=3}^{\mathrm{\infty }}\frac{1}{\left(k-3\right)!}+\sum _{k=2}^{\mathrm{\infty }}\frac{3}{\left(k-2\right)!}+\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\left(k-1\right)!}$ as $\frac{1}{\left(-r\right)!}=0$ for $r>0$ $=e+3e+e=5e$