Convergence of $\sum _{n=1}^{\infty }\ln\left(\frac{n^{-a}}{\sin(n^{-a})}\right)$ where $a>0$ I need help with this series. In order to meet the necessary convergence condition I know that $a$ has to be more than zero. I tried to use comparison test with $1/n^2$ but the limit seems to be infinity. The solution should be that it converges if $a>0.5$. Any suggestions?

$$\sum _{n=1}^{\infty }\ln\left(\frac{\frac{1}{n^a}}{\sin(\frac{1}{n^a})}\right), 
 \space a>0$$

Someone suggested use to Taylor... but i formally did not study it in school yet so  i would be grateful for hints without Taylor expansion.
 A: Note that as $x\to 0$,
$$\ln\left(\frac{x}{\sin(x)}\right)
=-\ln\left(\frac{\sin(x)}{x}\right)=-\ln\left(\frac{x-\frac{x^3}{6}+o(x^3)}{x}\right)\\=-\ln\left(1-\frac{x^2}{6}+o(x^2)\right)=\frac{x^2}{6}+o(x^2).$$
which implies that for $a>0$, 
$$\ln\left(\frac{\frac{1}{n^a}}{\sin(\frac{1}{n^a})}\right)\sim \frac{1}{6n^{2a}}.$$
Can you take it from here?
Edit. If Taylor expansion is not allowed, you may also evaluate the limit
$$\lim_{x\to 0}\frac{-\ln\left(\frac{\sin(x)}{x}\right)}{x^2}
=\lim_{x\to 0}\frac{\ln\left(1+\frac{\sin(x)-x}{x}\right)}{\frac{\sin(x)-x}{x}}\cdot \lim_{x\to 0}\frac{x-\sin(x)}{x^3}\\=\lim_{y\to 0}\frac{\ln\left(1+y\right)}{y}\cdot \lim_{x\to 0}\frac{1-\cos(x)}{3x^2}=1\cdot \lim_{x\to 0}\frac{\sin(x)}{6x}=\frac{1}{6}$$
by using Hopital's Theorem.
A: Note that for $x\to0$ we have
$$\csc{(x)}=\frac1x+\frac{x}6+o(x)$$
$$\ln{(1+x)}=x+o(x)$$
So our summand is asymptotic to
$$\begin{align}
\ln{\left(\frac1{n^a}\left(n^a+\frac1{6n^a}+o\left(\frac1{n^a}\right)\right)\right)}
&=\ln{\left(1+\frac1{6n^{2a}}+o\left(\frac1{n^{2a}}\right)\right)}\\
&=\frac1{6n^{2a}}+o\left(\frac1{n^{2a}}\right)\\
\end{align}$$
Then note that the series
$$\sum_{n=1}^\infty\left(\frac1{6n^{2a}}+o\left(\frac1{n^{2a}}\right)\right)$$
is a $p$-series and converges if and only if $a\gt\frac12$.
