Prove that functional series $\sum_{n=1}|\sin x|^{\sqrt{n}}$ is convergent when $x \in (-1,1)$ Prove that functional series $\sum_{n=1}|\sin x|^{\sqrt{n}}$ is convergent when $x \in (-1,1)$.
I do not know if that's so easy that I'm simply missing something, but I can't find any criterion which would solve this.
 A: We have $\vert\sin x\vert \le \vert x\vert$ and to conclude it suffices to prove that the series $\sum \vert x\vert^{\sqrt n}$ is convergent. We have
$$n^2 \vert x\vert^{\sqrt n}=\exp(\sqrt n\ln\vert x\vert+2\ln n)\xrightarrow{n\to\infty}0$$
so we have $\vert x\vert^{\sqrt n}\le \frac1{n^2}$ for $n$ large enough and the result follows by comparison with a Riemann convergent series.
A: Since $-1<x<1$, you have $|\sin x|<1$. Then $\log|\sin x|<0$. We have
$$
|\sin x|^{\sqrt n}=e^{\sqrt n\,\log|\sin x|}.
$$
If we now compare with the integral,
$$
\int_1^\infty e^{t\,\log|\sin x|}\,dt=\left.\frac{e^{t\,\log|\sin x|}}{\log|\sin x|}\right|_1^\infty=-\frac{e^{\log|\sin x|}}{\log|\sin x|}<\infty.
$$
So the series converges. 
A: $\sin(1)<0.9$ and $\sum\limits_{k=n^2+1}^{(n+1)^2} a^\sqrt{k}<(2n+1)a^n$.
$\sum\limits_{k=1}^\infty (\sin x)^\sqrt{k}\le \sum\limits_{k=1}^\infty (\sin 1)^\sqrt{k}< \sum\limits_{k=1}^\infty 0.9^\sqrt{k}\le \sum\limits_{k=1}^\infty (2k+1)0.9^k= 2\frac{0.9}{(1-0.9)^2}+\frac{0.9}{1-0.9}=189$ 
