Determine the convergence of the series log (n sin (1/n))

Consider the series

$\sum_{n=1}^{\infty} \log \left( n \sin \frac{1}{n} \right).$

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The given series converges absolutely.

Proof. First, we have

$n \sin \frac{1}{n} = \frac{\sin \frac{1}{n}}{\frac{1}{n}} < 1$

for all positive integers $n$ . (Since $\frac{\sin x}{x} < 1$ for $x > 0$ .) Therefore,

$\sum_{n=1}^{\infty}\left| \log \left( n \sin \frac{1}{n} \right)\right| = \sum_{n=1}^{\infty}-\log \left( n \sin \frac{1}{n} \right) = \sum_{n=1}^{\infty} \log \left( \frac{1}{n \sin \frac{1}{n}} \right).$

Then, we write the Taylor expansion of $\sin \frac{1}{n}$ ,

$\begin{align*} \sin \frac{1}{n} &= \frac{1}{n} - \frac{1}{6n^3} + \cdots \\[9pt] \implies n \sin \frac{1}{n} &= 1 - \frac{1}{6n^2} + \cdots \\[9pt] &\geq 1 - \frac{1}{6n^2} \\[9pt] &\geq 1 - \frac{1}{n^2}. \end{align*}$

Thus,

$\begin{align*} \sum_{n=2}^{\infty} \log \left( \frac{1}{n \sin \frac{1}{n}} \right) &\leq \sum_{n=2}^{\infty} \log \left( \frac{1}{1 - \frac{1}{n^2}} \right) \\[9pt] &= \sum_{n=2}^{\infty} \log \left( 1 - \frac{1}{n^2-1} \right). \end{align*}$

But,

$\log \left( 1 - \frac{1}{n^2-1} \right) < \frac{1}{n^2-1}$

for $n \geq 2$ since

$\begin{align*} e^{\frac{1}{n^2-1}} &= 1 + \left( \frac{1}{n^2-1} \right) + \cdots > 1 + \frac{1}{n^2-1}\\[9pt] \implies \frac{1}{n^2} &> \log \left( 1 + \frac{1}{n^2-1} \right). \end{align*}$