# Series $\frac{k^2}{k!}$ with $k=1$ to $\infty$ [duplicate]

A practice Math Subject GRE asked me to compute $\sum_{k=1}^\infty \frac{k^2}{k!}$. The sum is equal to $2e$, but I wasn't able to figure this out using Maclarin series or discrete PDFs. What's the most elementary way to solve this? Thank you.

• Are you familiar with Bell numbers? – Ali Aug 19 '13 at 12:56
• This is a duplicate (probably several times). – Did Aug 19 '13 at 15:12
• @Did I'm not surprised, but site and Google searches were insufficient to find duplicates. – tba Aug 19 '13 at 21:26

No theory needed; you can solve this with simple manipulations.

$$\sum_{1}^{\infty} \frac{k^2}{k!} = \sum_{1}^{\infty} \frac{k}{(k-1)!} = \sum_{0}^{\infty} \frac{k+1}{k!} = \sum_0^{\infty} \frac{1}{k!} + \sum_1^{\infty} \frac{k}{k!} = e + \sum_0^{\infty} \frac{1}{k!} = 2e$$

Consider the power series

$$f(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!} = e^x$$

$$\frac{d}{dx} \left [x \frac{d}{dx} f(x)\right] = \sum_{k=0}^{\infty} \frac{k^2 x^k}{k!} = (x+1) e^x$$

Plug in $x=1$ and the stated result follows.

I'll try to calculate the more general series: $$S_n=\sum_{k=0}^{\infty}\frac{k^n}{k!}$$ where $n$ is an integer number and $n\ge 0$.

\begin{align} \begin{aligned} S_n &=\sum_{k=0}^{\infty}\frac{k^n}{k!} \\ &=\sum_{k=1}^{\infty}\frac{k^n}{k!} \\ &= \sum_{k=1}^{\infty}\frac{((k-1)+1)^n}{k!}\\ &= \sum_{k=0}^{\infty}\frac{(k+1)^n}{(k+1)!} &k\to k+1\\ &= \sum_{k=0}^{\infty} \frac{(k+1)^{n-1}}{k!} &\text{simplifying}\\ &=\sum_{k=0}^{\infty}\frac{1}{k!}\left( \sum_{j=0}^{n-1}\binom{n-1}{j}k^j\right) &\color{green}{\text{binomial expansion}}\\ &= \sum_{j=0}^{n-1}\binom{n-1}{j}\sum_{k=0}^{\infty}\frac{k^j}{k!} &\color{red}{\text{rearranging}} \\ &= \sum_{j=0}^{n-1}\binom{n-1}{j} S_j \end{aligned} \end{align}

Now, if you are already familiar with Bell numbers, this will certainly ring a bell; otherwise it's a useful recursion relation to calculate the sum in the general case.

For $n=0$, by definition we know it's Euler's constant $S_0=e$. For $n=1$, we have:

$$S_1=\binom{0}{0}S_0=1 \times e = e$$

For $n=2$, which happens to be what has been asked, we have:

$$S_2=\binom 1 0 S_0 + \binom 1 1 S_1 = 2e$$

And so on. In general, since $S_n$'s have the same linear recursion relations as Bell numbers $B_n$, the solution would be $S_n=e B_n$.

A beautiful way of constructing these Bell numbers would be using the triangles which I've found in this post:

\begin{array}{ccccccccccccc} 1 & & 1 & & 2 & & \color{red}5 & & 15 & & 52 & & \color{blue}{203} \\ & 2 & & 3 & & 7 & & 20 & & 67 & & 255 \\ & & \color{red}{5} & & 10 & & 27 & & 87 & & 322 \\ & & & 15 & & 37 & & 114 & & 409 \\ & & & & 52 & & 151 & & 523 \\ & & & & & \color{blue}{203} & & 674 \\ & & & & & & 877 \end{array}

• @ABlumenthal I actually noticed the pun, after I made it $:)$. – Ali Aug 20 '13 at 1:43

$e^x=1+x+\dfrac{x^2}{2!}+\cdots+\dfrac{x^k}{k!}+\cdots$

Differentiating both sides,

$e^x=0+1+\dfrac{2x}{2!}+\cdots+\dfrac{kx^{k-1}}{k!}+\cdots$

Multiplying both sides by $x$,

$xe^x=x+\dfrac{2x^2}{2!}+\cdots+\dfrac{kx^k}{k!}+\cdots$

Again differentiating both sides,

$(x+1)e^x=1+\dfrac{2^2x}{2!}+\cdots+\dfrac{k^2x^{k-1}}{k!}+\cdots$

Now putting $x=1$ gives,

$2e=\dfrac{1^2}{1!}+\dfrac{2^2}{2!}+\cdots+\dfrac{k^2}{k!}+\cdots$