Radius of convergence of $\sum_{n=0}^\infty \frac{z^n}{e^{an}+1}, a \in \mathbb{C}$

Question

$$\sum_{n=0}^\infty \frac{z^n}{e^{an}+1}, a \in \mathbb{C}$$

I have some doubts on the result because of that complex parametre. I think that the radius is $R=0$, since the negative real axis of the complex plane makes the series diverge. Is it right?

Answer 1

Hint. For $a:=x+iy\in\mathbb{C}$, we have that $$|e^{an}+1|^{2}=|e^{xn}(\cos(ny)+i\sin(ny))+1|^{2}=e^{2xn}+2e^{xn}\cos(ny)+1.$$ Now evaluate the radius of convergence, $$\limsup_{n\to +\infty} |e^{an}+1|^{1/n}$$ by considering three cases: $x>0$, $x<0$, $x=0$ (harder).

Answer 2

No. The radius of convergence is at least $e^{\operatorname{Re}a}$, because$$|z|<\operatorname{Re}a\iff\left|\frac z{e^a}\right|<1\implies\sum_{n=0}^\infty\left(\frac z{e^a}\right)^n\frac1{1+e^{-an}}\text{ converges.}$$But$$\sum_{n=0}^\infty\left(\frac z{e^a}\right)^n\frac1{1+e^{-an}}=\sum_{n=0}^\infty\frac{z^n}{e^{an}+1}.$$