# Does the series $\sum\limits_{n=1}^{\infty}\frac{\sin(n-\sqrt{n^2+n})}{n}$ 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?

• Find an equivalent of the terms you sum. – Plop Nov 14 '10 at 17:38
• Although (some statements of) the comparison test requires positive terms, it can be combined with rules such as $$\sum_{n=1}^\infty c\,a_n=c\sum_{n=1}^\infty a_n$$ Setting $c=-1$ converts a negative series to a positive series. – robjohn Jan 12 '13 at 14:24

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.