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The sum to evaluate is: $$\sum_{n=0}^{\infty} \frac{1}{n! \, (n^4+n^2+1)}$$

As for what I've tried:

  • It is obvious that the $n$-th element converges to $0$
  • If we take $$\sum_{n=0}^\infty \frac{1}{n!}$$ and compare it to the sum we might get something but I hit a wall trying to do it.
Leucippus
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Boyan
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    "but I hit a wall trying to do it." Here is a tip for good question-asking (in life, but in particular on math.SE): if you say "I tried X and it didn't work," that provides almost nothing constructive for someone to respond. If you say, "Here is my attempt to do X, and this is where I got stuck" it provides the person helping you with actual information and lots of avenues to assist. – JBL Nov 22 '22 at 18:53
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    https://math.stackexchange.com/q/794272/42969, https://math.stackexchange.com/q/3339016/42969, https://math.stackexchange.com/q/1574311/42969 – all found with Approach0 – Martin R Nov 22 '22 at 18:57

1 Answers1

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Factor $(n^4 + n^2 +1) = (1 - n + n^2)(1 + n + n^2)$, break apart, note that $\sum\limits_{n=0}^\infty \frac{1}{n!} = e$, perform separate summations to find your answer is: $\frac{e}{2}$.

(I later found this is closely related to this solution, where the 1 comes from not having the $n=0$ term in the summation.)

David G. Stork
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