How to solve $\sum _{n=0}^{\infty }\:\frac{n^a}{n!}$ for any $a \geq 1$ in general?

Question

While doing a problem I encountered this summation

$$\sum _{n=1}^{\infty }\:\frac{n^2}{n!}=2e.$$

I used

$$\sum _{n=0}^{\infty }\:\frac{n}{n!}=e$$

and

$$\sum _{n=0}^{\infty }\:\frac{(n-1)^2}{n!}=e$$

these summations to obtain its value. Is there any more general way to do the problem, therefore a general solution for $\sum _{n=0}^{\infty }\:\frac{n^a}{n!}$ for any $ a \geq 1$?

Answer 1

After some playing around, I saw that it is connected to the Bell numbers. That wikipedia article also includes your sum: $$\frac{1}{e}\sum_{n=0}^\infty \frac{n^a}{n!} = B_a$$ where $B_a$ is the $a$-th Bell number. This equation is called Dobiński's formula. Take a look at its wikipedia page. The proof is not very easy.

Answer 2

Recall that the definition of the exponential function is given by

\begin{equation} e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}. \end{equation}

The trick now is to note that if we differentiate and multiply by $x$, we get

\begin{equation} x\frac{\mathrm{d}}{\mathrm{d}x}e^x=\sum_{n=0}^{\infty}\frac{n}{n!}x^n. \end{equation}

If we do this again, we get

\begin{equation} x\frac{\mathrm{d}}{\mathrm{d}x}\left(x\frac{\mathrm{d}}{\mathrm{d}x}e^x\right)=\sum_{n=0}^{\infty}\frac{n^2}{n!}x^n. \end{equation}

In full generality, define $D=x\frac{\mathrm{d}}{\mathrm{d}x}$ to be this differential operation. Then we have

\begin{equation} \sum_{n=0}^{\infty}\frac{n^a}{n!}=D^a(e^x)\bigg|_{x=1}. \end{equation}

Since $\frac{\mathrm{d}}{\mathrm{d}x}e^x$, it is easy to show inductively that

\begin{equation} D^a(e^x)=P_a(x)e^x \end{equation}

for some polynomial $P_a(x)$, in which case we can write

\begin{equation} \sum_{n=0}^{\infty}\frac{n^a}{n!}=P_a(1)e. \end{equation}

Now to actually find the polynomials $P_a(x)$. We can act with $D$ on $P_a(x)e^x$ to obtain

\begin{equation} P_{a+1}(x)e^x=D\left(P_a(x)e^x\right)=x\frac{\mathrm{d}P_a}{\mathrm{d}x}\,e^x+xP_a(x)e^x, \end{equation}

and so the polynomial $P_a(x)$ is recursively defined by

\begin{equation} P_{a+1}(x)=x\left(P_a(x)+\frac{\mathrm{d}P_a}{\mathrm{d}x}\right), \end{equation}

with $P_0(x)=1$. The first few of these polynomials are given by

\begin{equation} \begin{split} P_0(x)&=1,\\ P_1(x)&=x,\\ P_2(x)&=x^2+x,\\ P_3(x)&=x^3+3x+x,\\ P_4(x)&=x^4+6x^3+7x^2+x,\\ P_5(x)&=x^5+10x^4+25x^3+15x^2+x. \end{split} \end{equation}

One can inductively show that the polynomials $P_a(x)$ satisfy (this can be left as an exercise)

\begin{equation} P_{a+1}(x)=x\sum_{k=0}^{a}\binom{a}{k}P_{k}(x), \end{equation}

so if we define $B_n=P_n(1)$, then we have $B_0=1$ and

\begin{equation} B_{a+1}=\sum_{k=0}^{a}\binom{a}{k}B_k, \end{equation}

which is the recursive definition of the so-called Bell numbers. Thus, we have

\begin{equation} \sum_{n=0}^{\infty}\frac{n^a}{n!}=B_ae. \end{equation}