Exercise: Study the absolute convergence and convergence domains of the following series: $$\sum_\limits{n=1}^{\infty}(-1)^{n+1}e^{-n\sin x}$$
I tried to solve it in the following way:
My resolution:
Convergence: $\lim_{n\to\infty}(-1)^{n+1}e^{-n\sin x}=0$ So by the Leibniz criterion the series converge for any $x\in\mathbb{R}$.
Absolute convergence:$\sum_\limits{n=1}^{\infty}|(-1)^{n+1}e^{-n\sin x}|=\sum_\limits{n=1}^{\infty}|(-1)^{n+1}\frac{1}{e^{n\sin x}}|\leqslant\sum_\limits{n=1}^{\infty}|\frac{1}{e^{n\sin x}}|$ By the integral test $\int_\limits{1}^{\infty}\frac{1}{e^{n\sin x}} dn=\frac{1}{\sin x e^{\sin x}}<\infty$ So the series converge absolutely $\forall x\in\mathbb{R}$.
Solution(book): The series converge absolutely if $2k\pi<x<(2k+1)\pi\:\:\forall k\in \mathbb{Z}$, in all other points the series diverge.
Questions:
I was puzzled by the books solution. What have I done wrong?
How should I study the absolute convergence then?
Thanks in advance!
Note that when $x=\pi k$ then $\sin x=\sin \pi k=0$ for any $k\in\mathbb{N}$. Hence the series would be $$\sum_{n=1}^{\infty}(-1)^{n+1}e^{-n\sin x}=\sum_{n=1}^{\infty}(-1)^{n+1}e^{-n\sin \pi k}=\sum_{n=1}^{\infty}(-1)^{n+1}$$ which is not convergent.