Divergence of $ \sum_{n=1}^{\infty} \frac{\sin(\log n)}{n}$

Question

Can someone give me a hint on how to show $$ \sum_{n=1}^{\infty} \frac{\sin(\log n)}{n} = \sum_{n=1}^{\infty}\operatorname{Im} \frac{1}{n^{1+i}} $$ is divergent?

Answer 1

Hint (not a proof)

You know that the harmonic series diverges. Also $\ln n$ varies slowly when $n$ is growing. Therefore, you can find a long sequence of consecutive terms for which $\sin(\ln n) \ge 1/2$. The two arguments allow to find for all for all $N >0$, $q>p>N$ with

$$\sum_{k=p}^q \frac{\sin(\ln k)}{k} >1$$ in contradiction with Cauchy convergence criteria.