Answer
Divergent
Work Step by Step
Given $$\sum_{n=1}^{\infty}\frac{\sin (1 / n)}{\ln n}$$
Since for $n>2$
$$
\frac{\sin (1 / n)}{\ln n}>\frac{1}{2 n \ln n}
$$
Then
$$
\frac{\sin (1 / n)}{\ln n}>\frac{1}{2 n \ln n}
$$
By using the integral test for $\sum_{n=2}^{\infty} \frac{1}{2 n \ln n},$ consider $f(x)=\frac{1}{2 x \ln x}$
It is clear that $f(x)$ positive and decreasing, so
\begin{align*}
\int_{2}^{\infty} \frac{1}{2 x \ln x} d x&=\left.\frac{1}{2} \ln \ln x\right|_{2} ^{\infty}\\
=\infty
\end{align*}
Then $\sum_{n=2}^{\infty} \frac{1}{2 n \ln n}$ is divergent.
Hence $\sum_{n=2}^{\infty} \frac{\sin (1 / n)}{\ln n}$ is also divergent.
