<s>[INST] Does $\sum_{n=1}^{\infty}\ln(n\sin(\frac{1}{n}))$ converge? I must determine whether the following series converges: $$\sum_{n=1}^{\infty}\ln\left(n\sin\left(\frac{1}{n}\right)\right)$$
I know that in general, I must use the limit comparison test, but I cannot find an expression to which I can compare it. For instance, I have tried the usual process:
For $n$ large, we have that $\lim_{n\to \infty}n\sin\frac1n=1$, and so, $\ln(1)=0$. This fails the divergence test, but it cannot be concluded automatically that the series is convergent either. How may I proceed here? Any help would be appreciated.
 [/INST] Using Taylor series, $$\sin x\sim x-\frac{x^3}{3!}+o(x^3)$$$$\ln(1+x)\sim x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)$$
When $n\rightarrow\infty$, $$n\sin\frac{1}{n}\sim n(\frac{1}{n}-\frac{1}{3!n^3}+o(\frac{1}{n^3}))=1-\frac{1}{6n^2}+o(\frac{1}{n^2})$$
Thus, $$\ln(n\sin\frac{1}{n})\sim -\frac{1}{6n^2}+o(\frac{1}{n^2})-\frac{\left(-\frac{1}{6n^2}+o(\frac{1}{n^2})\right)^2}{2}+o\left(-\frac{1}{6n^2}+o(\frac{1}{n^2}\right)=-\frac{1}{6n^2}+o(\frac{1}{n^2})$$
As $\sum\dfrac{1}{n^2}$ is convergent, so is $\sum\ln(n\sin\dfrac{1}{n})$. 
Hope this can help you.
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