I'm interested in a method of evaluating $\sum_{n=0}^{\infty} \frac{n!}{(n^2)!}$.

If there was a linear equation with leading coefficient $1$ in the denominator or a quadratic with leading coefficient $1$ and a coefficient of $0$ on $x$ in the numerator I would've used partial fraction, or a method seen in this answer by Jack Evaluate $\frac{0!}{4!}+\frac{1!}{5!}+\frac{2!}{6!}+\frac{3!}{7!}+\frac{4!}{8!}+\cdots$. But those approaches don't seem to be working for this specific series.

Is it even possible to evaluate this to get a closed form?

  • $\begingroup$ Not really to me I have a factorial $(n^2)$, in that question there is no factorial of a quadratic. @Franky_GTH $\endgroup$ – Ahmed S. Attaalla Aug 7 '16 at 18:11
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    $\begingroup$ I doubt there is a closed form. However the convergence is very fast. $\endgroup$ – Marco Cantarini Aug 7 '16 at 18:17
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    $\begingroup$ Because the convergence is so rapid, the limit is apt to be a transcendental number. $\endgroup$ – hardmath Aug 7 '16 at 18:20
  • $\begingroup$ @hardmath: The convergence is certainly fast enough to show that the number is irrational. However, one would need each term to be smaller than the square of the previous term (roughly) to imply transcendence, and that isn't the case here. $\endgroup$ – Greg Martin Aug 7 '16 at 18:48
  • $\begingroup$ Some transformations can be made using Gauß' multiplication formula for $z=n$. $\endgroup$ – Rudi_Birnbaum Aug 7 '16 at 19:09

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