<s>[INST] Convergence of $\sum_{n=1}^{\infty} \log (n\sin\frac{1}{n})$ Convergence of $\sum_{n=1}^{\infty} \log (n\sin\frac{1}{n})$
First, we have
$$   n \sin \frac{1}{n} = \frac{\sin \frac{1}{n}}{\frac{1}{n}} < 1 $$
for all positive integers $n$. (Since $\frac{\sin x}{x} < 1 \forall x > 0$.) Therefore,
$$ \sum_{n=1}^{\infty}\left| \log \left( n \sin \frac{1}{n} \right)\right| = \sum_{n=1}^{\infty}-\log  \left( n \sin \frac{1}{n} \right) = \sum_{n=1}^{\infty} \log \left( \frac{1}{n \sin \frac{1}{n}} \right). $$
\begin{align*}  \sin \frac{1}{n} &= \frac{1}{n} - \frac{1}{6n^3} + \cdots \\[9pt]  \implies n \sin \frac{1}{n} &= 1 - \frac{1}{6n^2} + \cdots \\[9pt]  &\geq 1 - \frac{1}{6n^2} \\[9pt]  &\geq 1 - \frac{1}{n^2}. \end{align*}
Could someone please guide me on how to proceed from here?
Thanks a lot for your help.
 [/INST] We have that
$$\sin \frac{1}{n} = \frac{1}{n} - \frac{1}{6n^3} + O\left(\frac1{n^5}\right)$$
$$n\sin \frac{1}{n} = 1 - \frac{1}{6n^2} + O\left(\frac1{n^4}\right)$$
$$\log \left(n \sin \frac {1}{n}\right)=\log \left( 1 - \frac{1}{6n^2} + O\left(\frac1{n^4}\right)\right)=-\frac1{6n^2}+O\left(\frac1{n^4}\right)$$
therefore the given series converges by limit comparison test with $\sum \frac1{n^2}$.
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