<s>[INST] Calculate $\sum_{n=2}^\infty ({n^4+2n^3-3n^2-8n-3\over(n+2)!})$ Calculate $\sum_{n=2}^\infty ({n^4+2n^3-3n^2-8n-3\over(n+2)!})$
I thought about maybe breaking the polynomial in two different fractions in order to make the sum more manageable and reduce it to something similar to $\lim_{n\to\infty}(1+{1\over1!}+{1\over2!}+...+{1\over n!})$, but didn't manage
 [/INST] First step, we find the Taylor series of $x^4+2x^3-3x^2-8x-3$ at the point $x=-2$ and then use it to write
$$ n^4+2n^3-3n^2-8n-3 = 1-4\, \left( n+2 \right) +9\, \left( n+2 \right)^{2}-6\, \left( n+2
 \right) ^{3}+ \left( n+2 \right) ^{4}.$$
Using the above expansion and shifting the index of summation ($n \longleftrightarrow n-2$ ), we have 
$$ \sum_{n=2}^\infty {n^4+2n^3-3n^2-8n-3\over(n+2)!}= \sum_{n=2}^\infty {1-4\, \left( n+2 \right) +9\, \left( n+2 \right)^{2}-6\, \left( n+2\right) ^{3}+ \left( n+2 \right)^{4}\over(n+2)!} $$
$$ = \sum_{n=4}^\infty {1-4\,  n  +9\,  n^{2}-6\,  n^{3}+  n^{4} \over n! }+\sum_{n=0}^3 {1-4\,  n  +9\,  n^{2}-6\,  n^{3}+  n^{4} \over n! }$$
$$ -\sum_{n=0}^3 {1-4\,  n  +9\,  n^{2}-6\,  n^{3}+  n^{4} \over n! }$$ 
$$= c+ \sum_{n=0}^\infty {1-4\,  n  +9\,  n^{2}-6\,  n^{3}+  n^{4} \over n! } $$
$$ = c+e(1-4B_1 + 9 B_2 -6B_3 +B_4),  $$
where $B_n$ are the bell numbers
$$ B_n = \frac{1}{e}\sum_{k=0}^{\infty} \frac{k^n}{k!}, $$
and $c$ is given by
$$ c=-\sum_{n=0}^3 {1-4\,  n  +9\,  n^{2}-6\,  n^{3}+  n^{4} \over n! }. $$
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