<s>[INST] How to get the following summation of the series $\sum\limits_{n=0}^{\infty}\frac{1}{n!(n^4+n^2+1)}$ I am trying to find the sum $$\sum\limits_{n=0}^{\infty}\frac{1}{n!(n^4+n^2+1)}$$ I had factorized the sum as $$\frac{1}{2n(n!)}\left(\frac{1}{n(n-1)+1}-\frac{1}{n(n+1)+1}\right)$$ From this step, how to proceed?
 [/INST] Given your decomposition of the main term, summation by parts ensures that the series equals
$$1+\frac{1}{2}\sum_{n\geq 1}\left(\frac{1}{n\cdot n!}-\frac{1}{(n+1)\cdot(n+1)!}\right)\left(1-\frac{1}{n^2+n+1}\right)$$
or
$$ 1+\frac{1}{2}\sum_{n\geq 1}\frac{n^2+n+1}{n(n+1)(n+1)!}\cdot\frac{n(n+1)}{n^2+n+1}=1+\frac{1}{2}\sum_{n\geq 1}\frac{1}{(n+1)!}=1+\frac{1}{2}\sum_{m\geq 2}\frac{1}{m!}=1+\frac{e-2}{2}=\color{red}{\frac{e}{2}}.$$
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