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I have a question regarding the infinite series. I want to find a closed form expression for the following expression:

$$\sum_{n=1}^{+\infty}\frac{1}{\sqrt{n}}\frac{z^n}{n!}$$

where $z$ denotes the parameter. I don't know about the existence of any closed form expression or at least any approximation for the above series. I will be glad to see your comments and helps.

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  • $\begingroup$ Neither do I. Please let me know if you find one. $\endgroup$ – Lord Shark the Unknown Dec 2 '17 at 11:04
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    $\begingroup$ Related: math.stackexchange.com/questions/2117742/… $\endgroup$ – Gabriel Romon Dec 2 '17 at 11:08
  • $\begingroup$ @GabrielRomon Thanks alot! I studied the comments of your problem. However it seems that there are some mistake on the proposed solutions or at least I am not convinced with some of them. I am wondering how we could converge to a solution! $\endgroup$ – Ali Dec 2 '17 at 13:30
  • $\begingroup$ There are no simple closed forms, if not in terms of fractional Touchard polynomials. On the other hand the asymptotic behaviour for $x\to +\infty$ is pretty simple to study by squaring, see the question linked by Gabriel Romon. $\endgroup$ – Jack D'Aurizio Dec 2 '17 at 17:01
  • $\begingroup$ If you follow the advice tired gives in a comment to the OP in the other thread to express $\frac 1{\sqrt n}=\frac 1{\sqrt \pi}\int_0^\infty e^{-nq^2}dq$, you get $$\frac 1{\sqrt \pi}\int_0^\infty e^{ze^{-q^2}}dq$$ $\endgroup$ – Paul Sinclair Dec 2 '17 at 17:05

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