What is $ \sum_{n=0}^\infty \frac{2n^7 + n^6 + n^5 + 2n^2}{n!} $
Sum the infinite series $ \sum_{n=0}^\infty (2n^7 + n^6 + n^5 + 2n^2)/n! $
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0@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. – 2011-03-28
3 Answers
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 $e(2B_7 + B_6 + B_5 + 2B_2) = e(2 \times 877 + 203 + 52 + 2 \times 2) = 2013e$.
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0Sigh, I was aiming at 2011e and overshot via a typo. Cheers for the answer. – 2011-03-28
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 $z = 1$.