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Evaluate sum of the series: $\displaystyle \sum_{n=0}^\infty {1\over n!(n^4+n^2+1)}$

Comparison test $\left(\text {with } \sum {1\over n^2}\right)$ confirms the convergence of the series. But what's the sum of the series. Using comparison tests I managed to show that ${19\over 14}\le$ sum $\le{e+121\over 91}$.

The series seems to converge to $\frac e2$. But how to show that ?

Sam
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2 Answers2

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Let $ n\in\mathbb{N} $, notice that :\begin{aligned} \frac{1}{n^{4}+n^{2}+1}=\frac{1}{\left(n^{2}-n+1\right)\left(n^{2}+n+1\right)}&=\frac{1}{2}\left(\frac{n+1}{n^{2}+n+1}-\frac{n-1}{n^{2}-n+1}\right)\\ &=\frac{1}{2}\left(\frac{n+1}{n^{2}+n+1}-\frac{n^{2}}{n^{2}-n+1}+1\right) \end{aligned}

Thus : \begin{aligned} \small\sum_{n=0}^{+\infty}{\frac{1}{n!\left(n^{4}+n^{2}+1\right)}}&\small=\frac{1}{2}\sum_{n=0}^{+\infty}{\frac{1}{n!}\left(\frac{n+1}{n^{2}+n+1}-\frac{n^{2}}{n^{2}-n+1}+1\right)}\\ &\small=\frac{1}{2}\sum_{n=0}^{+\infty}{\left(\frac{\left(n+1\right)^{2}}{\left(n+1\right)!\left(\left(n+1\right)^{2}-\left(n+1\right)+1\right)}-\frac{n^{2}}{n!\left(n^{2}-n+1\right)}+\frac{1}{n!}\right)}\\&\small=\frac{1}{2}\sum_{n=0}^{+\infty}{\left(a_{n+1}-a_{n}\right)}+\frac{1}{2}\sum_{n=0}^{+\infty}{\frac{1}{n!}}\\ &\small=\frac{1}{2}\left(0-0\right)+\frac{\mathrm{e}}{2}\\&\small=\frac{\mathrm{e}}{2} \end{aligned}

Were $ \left(a_{n}\right)_{n\in\mathbb{N}} $ is the sequence defined by $ \left(\forall n\in\mathbb{N}\right),\ a_{n}=\frac{n^{2}}{n!\left(n^{2}-n+1\right)} $.

In the third line, we were able to split the sum apart, because $ \sum\limits_{n\geq 0}{\left(a_{n+1}-a_{n}\right)} $ and $ \sum\limits_{n\geq 0}{\frac{1}{n!}} $ both converges, which means $ \sum\limits_{n\geq 0}{\left(a_{n+1}-a_{n}+\frac{1}{n!}\right)} $ converges, and $ \sum\limits_{n=0}^{+\infty}{\left(a_{n+1}-a_{n}+\frac{1}{n!}\right)}=\sum\limits_{n=0}^{+\infty}{\left(a_{n+1}-a_{n}\right)}+\sum\limits_{n=0}^{+\infty}{\frac{1}{n!}} $.

CHAMSI
  • 8,333
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${1\over n!(n^4+n^2+1)}$ $=\frac 12 \left[{n\over (n+1)!\{n(n+1)n+1\}}-{n-1\over n!\{n(n-1)+1\}}+{1\over (n+1)!}\right]$

$\displaystyle \sum_{n=0}^\infty \textstyle {1\over n!(n^4+n^2+1)} =\displaystyle \frac 12 \left[ \sum_{n=1}^\infty \textstyle {n-1\over n! \{n(n-1)+1\}}-\displaystyle \sum_{n=0}^\infty \textstyle{n-1\over n!\{n(n-1)+1\}}+\displaystyle \sum_{n=0}^\infty \textstyle \frac{1}{n!}-1\right]$ $= \frac 12[-(-1)+e-1]=\frac e2$

Saikai Prime
  • 1,133