Simplification of a sum of factorials 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.
 A: 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 $z$. 

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 $\Gamma(z+1) / \Gamma(z^2+1)$
