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

Express $$n^4+2n^3-3n^2-8n-3=(n+2)(n+1)n(n-1)+B(n+2)(n+1)n+C(n+2)(n+1)+D(n+2)+E--->(1)$$
So that $$T_n=\frac{n^4+2n^3-3n^2-8n-3}{(n+2)!}=\frac1{(n-2)!}+\frac B{(n-1)!}+\frac C{(n)!}+\frac D{(n+1)!}+\frac E{(n+2)!}$$
Putting $n=-2$ in $(1), E=2^4+2\cdot2^3-3\cdot2^2-8\cdot2-3=1$
Similarly, putting  $n=-1,0,1$ we can find $D=0,C=-2,B=0$ .
$$\implies T_n=\frac{n^4+2n^3-3n^2-8n-3}{(n+2)!}=\frac1{(n-2)!}-\frac 2{n!} +\frac 1{(n+2)!}$$
Putting $n=2, T_2=\frac1{0!}-\frac 2{2!} +\frac 1{4!}$
Putting $n=3, T_3=\frac1{1!}-\frac 2{3!} +\frac 1{5!}$
Putting $n=4, T_4=\frac1{2!}-\frac 2{4!} +\frac 1{6!}$
$$\cdots$$
So, the sum will be $$\sum_{0\le r<\infty}\frac1{r!}-2\sum_{2\le s<\infty}\frac1{s!}+\sum_{4\le t<\infty}\frac1{t!}$$
$=\sum_{0\le r<\infty}\frac1{r!}-2\left(\sum_{0\le s<\infty}\frac1{s!}-\frac1{0!}-\frac1{1!}\right)+\sum_{0\le t<\infty}\frac1{t!}-\left(\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}\right)$
$$=e-2e+e-\{-2\left(\frac1{0!}+\frac1{1!}\right)+\left(\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}\right)\}=-\frac34$$
