Convergence of a Series involving Gamma Function

Question

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..

Answer 1

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.

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

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

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