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

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Homework Statement


Does the follow serie converge:
[tex]\sum_{n=1}^\infty sin(\frac{1}{n^2})[/tex]

Homework Equations


For serie [tex]a_n[/tex] and [tex]b_n[/tex] if:

A = [tex]0 \leq a_n \leq b_n[/tex]

if [tex]b_n[/tex] converges then [tex]a_n[/tex] converges

The Attempt at a Solution


I think that I have to use the equation (see 2) and then with

B = [tex]\sum_{n=1}^\infty \frac{1}{n^2}[/tex]

I think that it is larger than A. However I need proof... Any suggestions.

Thanks in advance.
 
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For x>=0, |sin(x)| <= x. (Better yet, on [0,1], 0 <= sin(x) <= x.) Or you can just use the limit comparison test.

Note that both convergence tests require your series to have nonnegative terms.
 
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