Problem_
Find the value of $$\sum_{n=0}^\infty \frac{1}{n!(n^4+n^2+1)}$$
It was quite difficult for me to deal with the given series because of the factorial in the denominator. For the series of factorial, I know the fact that: $$\sum_{n=0}^\infty{x^n\over n!}=e^x$$ Can we change the form of this series similar to(or exactly as same as) the given series?
When I segregate the series into partial sums, $$\sum_{n=0}^\infty \frac{1}{n!(n^4+n^2+1)}=\sum_{n=0}^\infty{1\over n!}{1\over2n}\left({1\over n^2-n+1}-{1\over n^2+n+1}\right)\\={1\over2}\sum_{n=0}^\infty{1\over (n+1)!-n!}\left({1\over n^2-n+1}-{1\over n^2+n+1}\right)$$
However, the terms are not eliminated by other terms, so I think it is not a right way. Besides, Riemann Sum did not give any any clues to solve. Is there other way to have the value of the series? If so, please teach me! Thanks.