calculus - Does the series $sumlimits_{n=1}^{infty}frac{sin(n-sqrt{n^2+n})}{n}$ converge?

I'm just reviewing for my exam tomorow looking at old exams, unfortunately I don't have solutions. Here is a question I found : determine if the series converges or diverges. If it converges find it's limit.

$$\displaystyle \sum\limits_{n=1}^{\infty}\dfrac{\sin(n-\sqrt{n^2+n})}{n}$$

I've ruled down possible tests to the limit comparison test, but I feel like I've made a mistake somewhere.
divergence test - limit is 0 by the squeeze theorem
integral test - who knows how to solve this
comparison test - series is not positive
ratio root tests - on the absolute value of the series, this wouldn't work out
alternating series test - would not work, the series is not decreasing or alternating

Any ideas what to compare this series here with or where my mistake is on my reasoning above?

Answer

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.