Calculus (3rd Edition)

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.