6
$\begingroup$

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?
$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.