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I was studying the Gamma Functions, and while looking for different series involving Gamma functions, I saw the following one:

For all $c \in [0,\infty)$ and $\epsilon \in (0,\infty)$, the series: $$\sum_{n=1}^{\infty} \dfrac{c^{n}}{\Gamma(n\epsilon)}$$ converges.

Well, as an immediate observation, we can trivially ignore the case $c =0$, and start looking for the case $c \in (0,\infty)$. What I have tried so far, is the conventional Ratio test and Rabe's test. Now, I think there is some kind of obvious trick which I may not know lying here.

So, please let me know if you have any kind of way out.

Thanks in Advance..

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    Oh yes!! I forgot to mention, I have considered the case of $n \epsilon \lt 1$ & $|c| \lt 1$, which are easy (as it seems to me) by Dirichlet's Test. But, I am clueless about the general scenario. – 2017-02-22
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    :| Have you not looked at my answer yet? – 2017-02-22
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    oops... Sorry!! I just saw it now. Very sorry. Thanks.. :-) – 2017-02-22

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Use Stirling approximation (limit comparison test) and the root test. It should come through rather obviously.

Also use $\Gamma(n\epsilon)=\frac{\Gamma(1+n\epsilon)}{n\epsilon}$.

It then converges for $c\in(-\infty,0)$ by absolute convergence.

$$\lim_{n\to\infty}\frac{\Gamma(n\epsilon)}{\frac1\epsilon\sqrt{\frac{2\pi}n}\left(\frac{n\epsilon}e\right)^n}=1$$

$$\lim_{n\to\infty}\sqrt[n]{\left|\frac{c^n}{\frac1\epsilon\sqrt{\frac{2\pi}n}\left(\frac{n\epsilon}e\right)^n}\right|}=\lim_{n\to\infty}\frac{|c|\cdot e\cdot\epsilon\cdot\sqrt[n]n}{\sqrt[n]{2\pi}n\epsilon}=0$$

Thus, it converges for all $c\in\mathbb R$.

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    @ Simply Beautiful Art, I know Stirling's approximation. Please let me know how you are writing the first limit. I mean how you are replacing the factorial, appearing in Stirling's formula ?? – 2017-02-22
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    @user92360 I replaced the factorial with the Gamma function (they are equivalent), using the following: $$\Gamma(1+n)=n!,\qquad\Gamma(n)=\frac{\Gamma(1+n)}n$$ – 2017-02-22
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    But, here "$n\epsilon$" may NOT be an integer.Right ?? So, you can't write $\Gamma (1+n) = n!$ . Isn't it ?? – 2017-02-22
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    Yuup got it. Had to check the section on Stirling's Formula of the Wikipedia page on Gamma Function. Thanks a ton... (Y) – 2017-02-22
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    Of course, but Stirling's approximation works for the gamma function at non-integer values as well. – 2017-02-22