Convergence & absolute convergence of a complex series

Question

Analyse convergence and absolute convergence of a complex series: $$\sum_{n=1}^{\infty} \frac{(-1)^n}{n+ie^n} $$

I have that $\sum_{n=1}^{\infty} |\frac{(-1)^n}{n+ie^n}|=\sum_{n=1}^{\infty} \frac{1}{|n+ie^n|}=\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2+e^{2n}}}$, and from this point I don't know how to continue. On the other hand using Cauchy's or de'Alembert criterion seems little pointless here, but maybe I don't see something here. Any help is much appreciated.

Answer 1

$\sqrt {n^{2}+e^{2n}} \geq \sqrt {e^{2n}}= {e^{n}}$ and $\sum \frac 1 {e^{n}}$ is a convergent geometric series.

Answer 2

$\frac{1}{\sqrt{n^2+e^{2n}}} \le \frac{1}{\sqrt{e^{2n}}}=(\frac{1}{e})^n.$

Can you proceed ?

Answer 3

Since$$\lim_{n\to\infty}\frac{\left\lvert\frac{(-1)^n}{n+ie^n}\right\rvert}{\frac1{e^n}}=1$$and since $\sum_{n=1}^\infty\frac1{e^n}$ converges, your series converges absolutely.