# Evaluate $\sum_{k=1}^\infty \frac{k^2}{(k-1)!}$.

Evaluate $\sum_{k=1}^\infty \frac{k^2}{(k-1)!}$

I sense the answer has some connection with $e$, but I don't know how it is. Please help. Thank you.

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You can use Bell numbers. See here. –  Mhenni Benghorbal Oct 27 '12 at 8:25

The most elementary calculation is probably this one:

\begin{align*} \sum_{k\ge 0}\frac{(k+1)^2}{k!}&=1+\sum_{k\ge 1}\frac{k^2+2k+1}{k!}\\ &=\color{red}{1}+\sum_{k\ge 1}\frac{k}{(k-1)!}+2\sum_{k\ge 1}\frac1{(k-1)!}+\color{red}{\sum_{k\ge 1}\frac1{k!}}\\ &=\sum_{k\ge 1}\frac{k-1+1}{(k-1)!}+2\sum_{k\ge 0}\frac1{k!}+\color{red}{\sum_{k\ge 0}\frac1{k!}}\\ &=\color{blue}{1}+\sum_{k\ge 2}\frac1{(k-2)!}+\color{blue}{\sum_{k\ge 2}\frac1{(k-1)!}}+3\sum_{k\ge 0}\frac1{k!}\\ &=\color{blue}{\sum_{k\ge 0}\frac1{k!}}+4\sum_{k\ge 0}\frac1{k!}\\ &=5\sum_{k\ge 0}\frac1{k!}\\ &=5e\;. \end{align*}

One can also make use of the identity $$x^n=\sum_k{n\brace k}x^{\underline k}\;,$$ where ${n\brace k}$ is a Stirling number of the second kind and $x^{\underline k}\triangleq x(x-1)(x-2)\dots(x-k+1)$ is a falling power. In particular,

\begin{align*} k^3&=\sum_{k=0}^3{3\brace i}k^i\\ &=0\cdot k^{\underline 0}+1\cdot k^{\underline 1}+3\cdot k^{\underline 2}+1\cdot k^{\underline 3}\\ &=k+3k(k-1)+k(k-1)(k-2)\;, \end{align*}

whence

\begin{align*} \frac{k^2}{(k-1)!}&=\frac{k^3}{k!}=\frac{k+3k(k-1)+k(k-1)(k-2)}{k!}\\ &=\frac1{(k-1)!}+\frac3{(k-2)!}+\frac1{(k-3)!}\;, \end{align*}

and summing over $k$ yields $5e$ as before.

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Let's start with $e^x=\sum_{k=1}^\infty \frac{x^{k-1}}{(k-1)!}$ then : $$x\,e^x=\sum_{k=1}^\infty \frac{x^k}{(k-1)!}$$ $$x(x\,e^x)'=x\,e^x+x^2\,e^x=\sum_{k=1}^\infty \frac{k\,x^k}{(k-1)!}$$ $$x(x(x\,e^x)')'=x(e^x+3x\,e^x+x^2\,e^x)=\sum_{k=1}^\infty \frac{k^2\,x^k}{(k-1)!}$$ Set $x=1$ to get : $$5\,e=\sum_{k=1}^\infty \frac{k^2}{(k-1)!}$$

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Very nice, this way seems the most clear to me. –  littleO Oct 27 '12 at 7:49
Thanks @littleO ! The trick of using the theta operator $\theta=x\frac d{dx}$ works pretty often (two times here). –  Raymond Manzoni Oct 27 '12 at 7:52
Very Nice! Very Nice! Magic trick. –  Babak S. Oct 27 '12 at 9:16
Thanks @Babak generating functions tricks are indeed very nice! –  Raymond Manzoni Oct 27 '12 at 9:29
Excellent! Very insightful. –  Thomas Oct 27 '12 at 13:14

For $\frac{P(n)}{(n-r)!},$ where $P(n)$ is a polynomial.

If the degree of $P(n)$ is $m>0,$ we can write $P(n)=A_0+A_1(n-r)+A_2(n-r)(n-r-1)+\cdots+A_m(n-r)(n-r-1)\cdots\{(n-r)-(m-1)\}$

Here $k^2=C+B(k-1)+A(k-1)(k-2)$

Putting $k=1$ in the above identity, $C=1$

$k=2,B+C=4\implies B=3$

$k=0\implies 2A-B+C=0,2A=B-C=3-1\implies A=1$

or comparing the coefficients of $k^2,A=1$

So, $$\frac{k^2}{(k-1)!}=\frac{(k-1)(k-2)+3(k-1)+1}{(k-1)!}=\frac1{(k-3)!}+\frac3{(k-2)!}+\frac1{(k-1)!}$$

$$\sum_{k=1}^\infty \frac{k^2}{(k-1)!}=\sum_{k=1}^\infty \left(\frac{(k-1)(k-2)+3(k-1)+1}{(k-1)!}\right)$$ $$=\sum_{k=3}^\infty \frac1{(k-3)!}+\sum_{k=2}^\infty \frac3{(k-2)!}+\sum_{k=1}^\infty \frac1{(k-1)!}$$ as $\frac1 {(-r)!}=0$ for $r>0$

$=e+3e+e=5e$

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How do you think to create the first line? –  ᴊ ᴀ s ᴏ ɴ Oct 27 '12 at 7:36
@jasoncube, this is the de facto way of solving $\frac{P(n)}{(n-r)!}$ where $P(n)$ is a polynomial of $n$. Here the highest power of $n$ is $2,$ so, there will be $2+1=3$ constants. –  lab bhattacharjee Oct 27 '12 at 7:41
Exactly the way I approached this. –  robjohn Oct 27 '12 at 7:43

I have provided a proof of the general case here. The proof provided shows

$$\sum_{k=0}^{\infty} \frac{k^n}{k!} = T_n \cdot e$$

where $n$ is the $n^{th}$ Bell number. Yours is the case $n = 3$. Just for completeness, the formula above is called Dobinski's formula.

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Let's finish it in one line

$$\sum_{k=1}^\infty \frac{(k-2)(k-1)+3(k-1)+1}{(k-1)!}=5e$$

Chris.

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