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

Question

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

Answer 1

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 $e(2B_7 + B_6 + B_5 + 2B_2) = e(2 \times 877 + 203 + 52 + 2 \times 2) = 2013e$.

Answer 2

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).