I want to show that $\sum\limits_{k=1}^n\sin(\frac{x}{k})$, where $x\neq0$, diverges.
WLOG $x>0$, then I would simply opt for the limit comparison test which delivers: $$ \lim\limits_{k\to\infty}\frac{\frac{1}{k}}{\sin(\frac{x}{k})} ~\underset{\text{L'Hospital}}{=}~\lim\limits_{k\to\infty}\frac{1}{x\cos(\frac{x}{k})}=\frac{1}{x}>0. $$
So by limit comparison test the series $\sum\limits_{k=1}^n\frac{1}{k}$ and $\sum\limits_{k=1}^n\sin(\frac{x}{k})$ have the same convergence properties. Hence, $\sum\limits_{k=1}^n\sin(\frac{x}{k})$ diverges.
However, I was wondering if there is another approach which doesn't use limit comparison test? Maybe there is some clever manipualtion of the summand $\sin(\frac{x}{k})$?