Find this sum $$\sum_{n=1}^{\infty}\dfrac{1}{n!(n^4+n^2+1)}$$
My try: since $$n^4+n^2+1=(n^2+1)^2-n^2=(n^2+n+1)(n^2-n+1)$$ then $$\sum_{n=1}^{\infty}\dfrac{1}{n!(n^4+n^2+1)}=\sum_{n=1}^{\infty}\dfrac{1}{2n\cdot n!}\left(\dfrac{1}{n(n-1)+1}-\dfrac{1}{n(n+1)+1}\right)$$ then I can't go further.
Let $S$ denote the sum to be evaluated. For every $n$, $$ \frac2{n^4+n^2+1}=\frac{n+1}{n(n+1)+1}-\frac{n-1}{(n-1)n+1}, $$ hence $$ 2S=\sum_{n\geqslant1}\frac{n+1}{n(n+1)+1}\frac1{n!}-\sum_{k\geqslant1}\frac{k-1}{(k-1)k+1}\frac1{k!}. $$ The RHS is almost a telescoping series. To wit, the change of variable $n=k-1$ in the last sum and the fact that its $k=1$ term is zero yield $$ 2S=\sum_{n\geqslant1}\frac{n+1}{n(n+1)+1}\frac1{n!}-\frac{n}{n(n+1)+1}\frac1{(n+1)!}, $$ that is, $$ 2S=\sum_{n\geqslant1}\frac{(n+1)^2-n}{n(n+1)+1}\frac1{(n+1)!}\stackrel{k=n+1}{=}\sum_{k\geqslant2}\frac1{k!}=\mathrm e-\frac1{0!}-\frac1{1!}, $$ and finally, $$ S=\frac12\mathrm e-1. $$
