# Does $\sum_{n=2}^{\infty} \frac{1}{\ln(n)}\cdot\frac{1}{\sin(1/n)}$ converge?

Does $\sum_{n=2}^{\infty} \frac{1}{\ln(n)}\cdot\frac{1}{\sin(1/n)}$ converge?

I thought about Dirichlet's test, I know that $\frac{1}{\ln(n)}\ge\frac{1}{\ln(n+1)}$ and $\lim_{n \rightarrow \infty}\frac{1}{\ln(n)}=0$, but how I can continue?

• Note that certain special functions (like $\sin$ and $\ln$) have their own commands. For instance type \sin in stead of sin. Same for $\ln$. Also \geq is the command for the $\geq$ sympbol (and \leq gives $\leq$). – gebruiker Nov 10 '17 at 19:23

It diverges trivially: $$\frac1{\ln n}\cdot\frac1{\sin\cfrac1n}\sim_\infty\frac1{\ln n}\cdot\frac1{\cfrac1n}=\frac n{\ln n}\xrightarrow[n\to\infty]{}+\infty$$
Hint: $\sin(1/n) \le 1/n.\,\,\,$