Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
    
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
up vote 16 down vote accepted

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.

share|cite|improve this answer
    
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
1  
@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.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.