# Sum the infinite series $\sum_{n=0}^\infty (2n^7 + n^6 + n^5 + 2n^2)/n!$

What is $$\sum_{n=0}^\infty \frac{2n^7 + n^6 + n^5 + 2n^2}{n!}$$

• Note that at least one of the two respondents did not realize that you already knew (or thought you knew) the answer to the question before you asked it. It seems that just having the "puzzle" tag is not (yet) clear enough for that purpose. I would appreciate it if in the future you would make it more clear when you are posing a problem rather than asking a question by calling attention to this in the question itself (or perhaps even the title). – Pete L. Clark Mar 28 '11 at 5:53
• Can you suggest some words that would clearly and politely indicate this? Saying, "By the way, I know the answer" seems awkward; I was hoping that using the "puzzle" tag was sufficient. – Fixee Mar 28 '11 at 5:57
• Just a curiosity: I wrote a puzzle post some days ago, and it was closed in a flash... How comes this is still open? – Pacciu Mar 28 '11 at 9:02
• @Pacciu: I don't know (FWIW, I didn't vote to close your question). You may want to bring this up on the meta site. – Pete L. Clark Mar 28 '11 at 9:27

From Dobiński's formula, $\sum \limits_{n=0}^\infty \frac{n^k}{n!} = eB_k$, where $B_k$ is the $k$th Bell number, the number of set partitions of a set of size $k$..
So the answer is $e(2B_7 + B_6 + B_5 + 2B_2) = e(2 \times 877 + 203 + 52 + 2 \times 2) = 2013e$.
Here's a hint. $$\sum_{n=0}^{\infty} \frac{n^k}{n!}=e B_k$$ where $B_k$ is the $k$-th Bell number (see Dobiński's formula for a proof).
You have that if: $$A(z) = \sum_{n \ge 0} a_n z^n$$ then: $$z \frac{\mathrm{d}}{\mathrm{d} z} A(z) = \sum_{n \ge 0} n a_n z^n$$ In general, using $\mathrm{D}$ for the differentiation operator, and $p$ a polyomial: $$\sum_{n \ge 0} p(n) a_n z^n = p(z \mathrm{D}) A(z)$$ So you are looking for: $$(2 (z \mathrm{D})^7 + (z \mathrm{D})^6 + (z \mathrm{D})^5 + 2 (z \mathrm{D})^2) \mathrm{e}^z$$ at $z = 1$.