# Convergence of $\sum_{n=1}^\infty \frac{\sin^2(n)}{n}$

Does the series $$\sum_{n=1}^\infty \frac{\sin^2(n)}{n}$$ converge?

I've tried to apply some tests, and I don't know how to bound the general term, so I must have missed something. Thanks in advance.

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Why not asking this on the page of your previous question? – Did Jan 9 '13 at 16:07
@did Now I can't delete this question. I'm sorry. Thanks for this example. – V. Galerkin Jan 9 '13 at 16:57

## 2 Answers

Hint:

Each interval of the form $\bigl[k\pi+{\pi\over6}, (k+1)\pi-{\pi\over6}\bigr)$ contains an integer $n_k$. We then have, for each $k$, that ${\sin^2(n_k)\over n_k}\ge {(1/2)^2\over (k+1)\pi}$. Now use a comparison test to show your series diverges.

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How could you prove that in every interval $\bigl[k\pi+{\pi\over6}, (k+1)\pi-{\pi\over6}\bigr)$ has an integer $n_k$? – Zack Ni Jun 21 at 13:05
@ZackNi The length of such an interval is $\pi-\pi/3>1$. – David Mitra Jun 21 at 13:10

It is divergent:

Write $$\sum \frac{\sin^2(n)}{n} = \sum \frac{1}{2n} - \sum \frac{\cos(2n)}{2n},$$ clearly at the right handed side, the first is divergent but the second converges.

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