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I am having trouble simplifying the following series. The sum is as follows: $$\sum_{k=0}^{\infty} \frac{x^k}{k^2!}.$$ is it possible to simplify? My intuition is telling me that a simplification involving the exponential is possible. Help would be appreciated, thank you.

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    $\begingroup$ $\sum x^k / (k!)^2$ is essentially a Bessel function. But I think $k^2!$ is not found in any named function. $\endgroup$ – GEdgar Jan 29 '20 at 1:05
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    $\begingroup$ One could write a contour integral, say $$\sum_{k=0}^\infty\frac{x^k}{k^2!}=\frac{1}{2\pi i}\oint_{|z|=r}\frac{e^z}{z}\sum_{k=0}^\infty\frac{x^k}{z^{k^2}}\,dz\qquad(r>1)$$ where some sort of theta function appears. Of course there's a real version of this integral. But... this is not of too much help, and I wouldn't expect to see any. $\endgroup$ – metamorphy Mar 19 '20 at 8:31
  • $\begingroup$ Supposing that evaluating this series at zero is infinitely differentiable, then we can rewrite the series as a Maclaurin series, showing us $f^{\left(k\right)}\left(0\right)=\frac{k!}{\left(k+1\right)^{2}!},\ \forall k\in\mathbb{W}$. This would also show $f(0)=1$. But given this information, I doubt it could be used beneficially, since we still retain that nasty term of the factorial of a square. $\endgroup$ – Void Mar 14 at 2:53
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I think I have a partial answer. Begin by recognizing the square factorial function can be broken up into two components:

$$ k^2 ! = k! \prod_{j=k+1}^{k^2} j $$ The series is now: $$ \sum_{k=0}^\infty \frac{x^k}{k!} \prod_{j=k+1}^{k^2} \frac{1}{j} $$ The product can be expressed as $$ \frac{\Gamma(k+1)}{\Gamma(k^2+1)} = \frac{\Gamma(k)}{k\Gamma(k^2)} $$ The second equality is obtained by $z\Gamma(z) = \Gamma(z+1)$ (alternatively, $\Gamma(z+1) = \Pi(z)$. Hence the series can be written as:

$$ \sum_{k=0}^\infty \frac{x^k}{k!} \frac{\Gamma(k)}{k\Gamma(k^2)} = \sum_{k=0}^\infty \frac{x^k}{k!} \frac{\Pi(k)}{\Pi(k^2)} $$

I have looked through many references and I cannot find an identity that simplifies the ratio of the Pi or Gamma functions. Unfortunately duty calls and I cannot spend anymore time on this problem, but I do believe this is the best that can possibly be done. However in this form it is easy to ascertain the following:

1) The series converges (to what, we still don't know)

2) The converged value must be less than $e^x $, because the ratio of $\Pi(z)/ \Pi(z^2) \leq 1$ for most of $z$.

Ratio of Gamma(z squared) to Gamma(z squared +1)

For the hell of it I checked Wolfram's infinite series calculator. I was only told the series converges (as expected), but no specific answer was given. Of course, these types of calculators aren't 100% reliable and dont replace a proof. If anyone has further suggestions please chime in. For whatever reason the caption on the picture isnt adding nicely. This is a graph of $\Gamma(z+1) / \Gamma(z^2+1)$

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    $\begingroup$ I don't think there is a closed form expression. In my strategy I was trying to convert the series into something more familiar. In this case it seems to be the terms of the Taylor Series of e^x multiplied by the ratio of Pi(k)/Pi(k^2). In this case you can show it converges and is less than e^x, but unless there is a simplification of that ratio term I don't think it's possible to go further, unfortunately. There are some esoteric identities, but none that are applicable that I saw. $\endgroup$ – Alex Butkus Mar 25 '20 at 4:54

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