Does the series $\sum\limits_{n=1}^{\infty}\frac{\sin(n-\sqrt{n^2+n})}{n}$ converge?
The key here is that $n - \sqrt{n^2 + n}$ converges to $-{1 \over 2}$ as $n$ goes to infinity: $$n - \sqrt{n^2 + n}= (n - \sqrt{n^2 + n}) \times {n + \sqrt{n^2 + n} \over n + \sqrt{n^2 + n}}$$ $$= {n^2 - (n^2 + n) \over n + \sqrt{n^2 + n}} = -{n \over n + \sqrt{n^2 + n}}$$ $$= -{1 \over 1 + \sqrt{1 + {1 \over n}}}$$ Take limits as $n$ goes to infinity to get $-{1 \over 2}$.
Hence $\sin(n - \sqrt{n^2 + n})$ converges to $\sin(-{1 \over 2})$, and the series diverges similarly to ${1 \over n}$, using the limit comparison test for example.