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How would you find the sum of the following series. $\frac{k^3+6k^2+11k+5}{(k+3)!}$ as k goes from 1 to infinity

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    That is an expression, not a series. Do you mean for that to be the summand? from what value to what value? A suggestion would be to rename variables and write it so that you have an $n!$ in the denominator and have it as $\sum\frac{a_1 n^3}{n!} + \sum\frac{a_2 n^2}{n!}+\sum\frac{a_3n}{n!}+\sum\frac{a_4}{n!}$ for constants $a_1,a_2,a_3,a_4$, each of which should be a known form – 2017-01-16

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Hint you can see that it can be weitten as $$\frac {(k+1)(k+2)(k+3)}{(k+3)!}-\frac{1 }{(k+3)!} $$ now thats equal to $\frac {1}{k!}-\frac {1}{(k+3)!} $ also note that $\sum _0 ^\infty \frac {1}{n!}=e $ thus you can now find the answer

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In a general way, for the series $$\sum_{n=1}^{+\infty}\frac{P(n)}{(n+a)!}\text{ with }a\in \mathbb{N} \text{ and }P\text{ polynomial of degree }k,$$ we can express the numerator $$\begin{aligned}&P(n)=A_k\underbrace{(n+a)(n+a-1)\ldots}_{k\text{ factors}}+A_{k-1}\underbrace{(n+a)(n+a-1)\ldots}_{k-1\text{ factors}}\\&+\cdots+A_2\underbrace{(n+a)(n+a-1)}_{2\text{ factors}}+A_1\underbrace{(n+a)}_{1\text{ factor}}+A_0,\end{aligned}$$ symplify, decompose in sum of series and use $e=\sum_{m=0}^{+\infty}\frac{1}{m!}.$