# Exploring $\sum_{n=0}^\infty \frac{n^p}{n!} = B_pe$, particularly $p = 2$.

I was exploring the fact that $$\sum_{n=0}^\infty \frac{n^p}{n!} = B_pe,$$ where $B_n$ is the $n$th Bell number.

I found this result by exploring the series on wolframalpha and looking up the sequence of numbers generated. I have no experience with Bell numbers other than knowing that they represent the number of ways of partitioning a set.

What I tried

Looking at the base case we can verify $$\sum_{n=0}^\infty \frac{n}{n!} = \sum_{n=1}^\infty \frac{1}{(n-1)!} = e.$$

But then I realize that $n^2$ is going to be difficult.

Specifically what I want

I am doing this for fun, so I just want to glean some sort of lesson out of this. Resources that let me learn for myself are just as good!

An accepted answer will have any one of the following:

• A proof that $\sum_{n = 0}^{\infty} \frac{n^2}{n!} = 2e$ (this interests me a lot)
• A description of why this series is related to Bell numbers and the number of partitions of a set
• A link to some resource that introduces Bell numbers and/or their connection to this series.
• Hi Chantry. Your first two questions are completely answered below. – Berrick Caleb Fillmore Apr 20 '15 at 4:22
• Hint: Since $n!~=~1\cdot2\cdot3\cdots(n-1)~n$, an obvious approach would be rewriting $n^2$ as $n~(n-1)+n$. – Lucian Apr 20 '15 at 9:19

Notice that \begin{align} S & = \sum_{n = 0}^{\infty} \frac{n^{2}}{n!} \\ & = \frac{0^{2}}{0!} + \frac{1^{2}}{1!} + \frac{2^{2}}{2!} + \frac{3^{2}}{3!} + \cdots \\ & = \frac{1}{0!} + \frac{2}{1!} + \frac{3}{2!} + \frac{4}{3!} + \cdots. \end{align} Then \begin{align} S - e & = \left( \frac{1}{0!} + \frac{2}{1!} + \frac{3}{2!} + \frac{4}{3!} + \cdots \right) - \left( \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots \right) \\ & = \frac{1}{1!} + \frac{2}{2!} + \frac{3}{3!} + \cdots \\ & = e. \qquad (\text{As shown in the OP.}) \\ \end{align} Therefore, $S = e + e = 2 e$.

In fact, using the same type of reasoning, it can be shown that $\displaystyle \sum_{n = 0}^{\infty} \frac{n^{3}}{n!} = 5 e$.

Here is the connection with Bell numbers.

For each $p \in \mathbb{N}_{0}$, let $\displaystyle S_{p} \stackrel{\text{df}}{=} \sum_{n = 0}^{\infty} \frac{n^{p}}{n!}$. Notice that \begin{align} \forall p \in \mathbb{N}: \quad S_{p} & = \sum_{n = 0}^{\infty} \frac{n^{p}}{n!} \\ & = \sum_{n = 1}^{\infty} \frac{n^{p}}{n!} \qquad \left( \text{As $\dfrac{0^{p}}{0!} = 0$.} \right) \\ & = \sum_{n = 1}^{\infty} \frac{n^{p - 1}}{(n - 1)!} \qquad (\text{After canceling $n$’s.}) \\ & = \sum_{n = 0}^{\infty} \frac{(n + 1)^{p - 1}}{n!}. \qquad (\text{After re-indexing.}) \end{align} Hence, \begin{align} \forall p \in \mathbb{N}_{\geq 2}: \quad S_{p} - S_{p - 1} & = \sum_{n = 0}^{\infty} \frac{(n + 1)^{p - 1}}{n!} - \sum_{n = 0}^{\infty} \frac{n^{p - 1}}{n!} \\ & = \sum_{n = 0}^{\infty} \frac{1}{n!} \left[ (n + 1)^{p - 1} - n^{p - 1} \right] \\ & = \sum_{n = 0}^{\infty} \left[ \frac{1}{n!} \sum_{k = 0}^{p - 2} \binom{p - 1}{k} n^{k} \right] \qquad (\text{By the Binomial Theorem.}) \\ & = \sum_{n = 0}^{\infty} \sum_{k = 0}^{p - 2} \binom{p - 1}{k} \frac{n^{k}}{n!} \\ & = \sum_{k = 0}^{p - 2} \left[ \binom{p - 1}{k} \sum_{n = 0}^{\infty} \frac{n^{k}}{n!} \right] \\ & = \sum_{k = 0}^{p - 2} \binom{p - 1}{k} S_{k}. \end{align} Therefore, $$\forall p \in \mathbb{N}_{\geq 2}: \quad S_{p} = \sum_{k = 0}^{p - 2} \binom{p - 1}{k} S_{k} + S_{p - 1} = \sum_{k = 0}^{p - 1} \binom{p - 1}{k} S_{k}.$$ As $S_{0} = S_{1} = e$ by inspection, we see that the $S_{p}$’s are the Bell numbers multiplied by $e$.

• Wonderful! This looks like it extends well to higher cases too - we can use that $\sum_{n=0}^{\infty}\frac{n^p}{n!}=\sum_{n=0}^{\infty}\frac{(n+1)^{p-1}}{n!}$ and then expand the $(n+1)^{p-1}$ term to get that sum in terms of various sums of the form $\sum_{n=0}^{\infty}\frac{n^{p'}}{n!}$ in terms of lesser $p'$ (which implies some recurrence relation I don't feel like writing out). – Milo Brandt Apr 20 '15 at 3:35
• @Meelo: Yes, there exists a recurrence relation. I once had to derive it in a high-school exam. – Berrick Caleb Fillmore Apr 20 '15 at 3:37

Recall the species of set partitions with the constituents marked which is $$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$ which gives the generating function $$G(z, u) = \exp(u(\exp(z)-1)).$$

It follows that the exponential generating function of Bell numbers is given by $$G(z) = \exp(\exp(z)-1).$$

Suppose we are trying to compute $$\sum_{n\ge 0} \frac{n^p}{n!}.$$

Put $$n^p = \frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} \exp(nz) \; dz.$$

Observe that this gives $n^p = 0$ when $n=0$ except when $p=0$ as well.

We get for the sum $$\frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} \sum_{n\ge 0} \frac{\exp(nz)}{n!} \; dz = \frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} \exp(\exp(z)) \; dz \\ = \exp(1) \times \frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} \exp(\exp(z)-1) \; dz \\ = \exp(1)\times B_p$$ as claimed.

• This is a little beyond my current understanding. I do not know what $\mathfrak{P}$ (power set maybe?) stands for and have little understanding about generating functions and contour integrals. Thank you for your contribution though! I will be sure to return to your answer in the future after studying both subjects. – Chantry Cargill Apr 21 '15 at 0:38