TITLE: Does the series $\sum\limits_{n=1}^{\infty}\frac{\sin(n-\sqrt{n^2+n})}{n}$ converge?
QUESTION [32 upvotes]: 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?

REPLY [39 votes]: 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.