<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 $\sin \dfrac{1}{n} = \dfrac{1}{n}-\dfrac{1}{3! n^3} + o\left(\dfrac{1}{n^3}\right)$, then :
$$\log\left( n \sin \dfrac{1}{n} \right) = \log\left( 1-\dfrac{1}{3! n^2} + o\left(\dfrac{1}{n^2}\right) \right) = -\dfrac{1}{3! n^2} + o\left(\dfrac{1}{n^2}\right) + \mathcal{O}\left( \dfrac{1}{n^4} \right)$$
Then : 
$$\log\left( n \sin \dfrac{1}{n} \right) = \left(-\dfrac{1}{6}+o(1)\right) \dfrac{1}{n^2}+ \mathcal{O}\left( \dfrac{1}{n^4}\right)$$
Then: 
$$n \to +\infty, \log\left( n \sin \dfrac{1}{n} \right) \sim \dfrac{-1}{6n^2}$$
</s>