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I am wondering how to evaluate the following sum:

$$\sum_{n=0}^{\infty}\frac{1}{(n^4+n^2+1)n!}.$$

In wolfram alpha I find it is equal to $e/2$ .

I have used the residue method but I didn't succeed and also using digamma function is still hard for me, my problem is treating the $n!$.

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  • $\begingroup$ i would try to use a partial fraction decomposition, followed by replacing $(n-n_i)^{-1}=-\int_0^{\infty} dt e^{-(n-n_i)t}$, where $n_i$ is a root of the polynomial in the denoinator. The summation becomes almost trivial and the resulting integral doesn't look undoable... $\endgroup$ – tired Dec 13 '15 at 23:28
  • $\begingroup$ I look to use taylor expansion of \pi cot(\piz) to do transformation using resudue method bu what about n! ? $\endgroup$ – Salmahamizi Hamizi Dec 13 '15 at 23:34
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We can rewrite the prefactor as $$\frac{1}{n^4+n^2+1}=\frac{1}{(n^2-n+1)(n^2+n+1)}=a_{n+1}-na_n+\frac12,$$ with $\displaystyle a_n=\frac{n}{2(n^2-n+1)}$. Now it is easy to understand that $a_n$'s give a sum that telescopes to $0$, so that we are left with $$\frac12\sum_{n=0}^{\infty}\frac{1}{n!}=\frac e2.$$


Added on request of OP: $$\sum_{n=0}^{\infty}\frac{-na_n+a_{n+1}}{n!}=-\frac{0\cdot a_0}{0!}+{\color{red}{\frac{a_{1}}{0!}-\frac{1\cdot a_1}{1!}}}+{\color{blue}{\frac{a_2}{1!}-\frac{2\cdot a_2}{2!}}}+{\color{magenta}{\frac{a_3}{2!}-\frac{3\cdot a_3}{3!}}}+\frac{a_4}{3!}+\ldots=0.$$

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  • $\begingroup$ @start wearing purple , could u show me a little bit how a_n give a sum that telescopes to 0 , i have checked it out but it's not work ? $\endgroup$ – Salmahamizi Hamizi Dec 14 '15 at 22:21
  • $\begingroup$ @Start wearing purple , I have checked to look the telescoping of a_n to 0 but it's not work and pleas check :wolframalpha.com/input/?i=+\sum%28\frac{n}{%28n%C2%B2-n%2B1%29n!}+%2Cn%3D+0+to+infty+ $\endgroup$ – Salmahamizi Hamizi Dec 15 '15 at 22:24
  • $\begingroup$ @SalmahamiziHamizi Is it more clear now? Your link does not work for me, however. $\endgroup$ – Start wearing purple Dec 16 '15 at 13:45

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