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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.

StephenG - Help Ukraine
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Ali
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  • Neither do I. Please let me know if you find one. – Angina Seng Dec 02 '17 at 11:04
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    Related: https://math.stackexchange.com/questions/2117742/lim-x-rightarrow-infty-sqrtxe-x-left-sum-k%EF%BC%9D1-infty-fracxk – Gabriel Romon Dec 02 '17 at 11:08
  • @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! – Ali Dec 02 '17 at 13:30
  • 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. – Jack D'Aurizio Dec 02 '17 at 17:01
  • 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 $$\frac 1{\sqrt \pi}\int_0^\infty e^{ze^{-q^2}}dq$$ – Paul Sinclair Dec 02 '17 at 17:05
  • Thank you for your precise comment. I came off with the above expression you mentioned (the integral), however I could not find any closed form for it in the handbooks of special integrals and functions. Is this an open problem in the mathematics literature? If no, how could I solve the above integral or at least find a more simple approximation for it? @JackD'Aurizio – Ali Dec 03 '17 at 09:26

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