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?
How to get the following summation of the series $\sum\limits_{n=0}^{\infty}\frac{1}{n!(n^4+n^2+1)}$
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1Must there an answer or you are just suddenly thought of? – MafPrivate Aug 30 '19 at 11:43
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@IsaacYIUMathStudio The sum results in $e/2$. See here. – Peter Foreman Aug 30 '19 at 11:45
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The answer is $\dfrac{e}{2}$ – vqw7Ad Aug 30 '19 at 12:03
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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 $$ 1+\frac{1}{2}\sum_{n\geq 1}\frac{n^2+n+1}{n(n+1)(n+1)!}\cdot\frac{n(n+1)}{n^2+n+1}=1+\frac{1}{2}\sum_{n\geq 1}\frac{1}{(n+1)!}=1+\frac{1}{2}\sum_{m\geq 2}\frac{1}{m!}=1+\frac{e-2}{2}=\color{red}{\frac{e}{2}}.$$
Jack D'Aurizio
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@J.Doe: summation by parts gives that $$\sum_{n=1}^{N}a_n b_n = A_N b_N - \sum_{n=1}^{N-1}A_n (b_{n+1}-b_n)$$ where $A_n=a_1+a_2+\ldots+a_n$. In our case we take $a_n=\frac{1}{n(n-1)+1}-\frac{1}{(n+1)n+1}$, $b_n=\frac{1}{2n\cdot n!}$, then consider the limit as $N\to +\infty$. – Jack D'Aurizio Aug 30 '19 at 13:13
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