While working on this question I think I've found a closed-form expression for the following series, but I don't know how to prove it.

Let $a \in \mathbb{N}$ and $b \in \mathbb{R}$. Then

$$\sum_{k=0}^{\infty} \frac{k^a\,b^k}{k!} \stackrel{?}{=} e^b \sum_{j=0}^a S(a,a-j+1)\,b^{a-j+1}, \tag{1}$$

where $e$ is Euler's number and $S(n,k)$ are the Stirling numbers of the second kind, defined as

$$S(n,k) = \frac{1}{k!}\sum_{i=0}^k (-1)^{i}{k \choose i} (k-i)^n,$$

with ${n \choose k}$ a binomial coefficient.

I have three questions.

  • $1^\text{st}$ Question. Is $(1)$ true?
  • $2^\text{nd}$ Question. If $(1)$ is true, then can we write $(1)$ into a more compact form with solving the finite sum somehow?
  • $3^\text{rd}$ Question. If $(1)$ is true, then can we generalize the statment for $a \in \mathbb{R}$? I don't know about this kind of generalization of the Stirling numbers of the second kind. Maybe there is another approach?
  • $\begingroup$ Hope it will be helpful: if $a=12$ and $b=\pi$, then, up to Maple, the sum of the series under consideration equals $$\pi \,{{\rm e}^{\pi }} \left( {\pi }^{11}+66\,{\pi }^{10}+1705\,{\pi } ^{9}+22275\,{\pi }^{8}+159027\,{\pi }^{7}+627396\,{\pi }^{6}+1323652\, {\pi }^{5}+1379400\,{\pi }^{4}+611501\,{\pi }^{3}+86526\,{\pi }^{2}+ 2047\,\pi +1 \right) $$ . $\endgroup$ – user64494 Oct 18 '14 at 20:09

Notice that

$$x^m e^x = \sum_{k=0}^\infty \frac{1}{k!} x^{m+k} = \sum_{j = m}^\infty \frac{1}{(j-m)!} x^j$$

So we have

\begin{align*} \sum_{k=0}^\infty \frac{k^m}{k!} x^k &= \sum_{k=0}^{m-1} \frac{k^m}{k!} x^k + \sum_{k=m}^\infty \frac{k^m}{k!} x^k \\ &= \sum_{k=0}^{m-1} \frac{k^m}{k!} x^k + \sum_{k=m}^\infty \frac{1}{(k-m)!} x^k \\ &\qquad \qquad+ \sum_{k=m}^\infty \frac{k^m - k(k-1)\cdots(k-m+1)}{k!} x^k \\ &= x^m e^x + \sum_{k=0}^{m-1} \frac{k^m}{k!} x^k +\sum_{k=m}^\infty \frac{k^m - k(k-1)\cdots(k-m+1)}{k!} x^k \\ \end{align*}

Inducing on $m$, you should be able to write the remaining sums in terms of things you already know.


The closed form is $e^b \sum\limits_{i = 0}^a S(a, i) b^i$.

Let $F(a, b)$ be the problem sum. Notice that $F(0, b) = e^b$ and $F(a, b) = b\frac{\partial F(a-1,b)}{\partial b}$. Therefore $F(a, b) = e^bP_a(b)$ for some polynomial $P_a$. It only remains to observe that the coefficients of $P_a$ satisfy the same recurrence as Stirling numbers of the second kind.


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