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Convergence sum sin(1/n^2)

  1. Oct 8, 2007 #1
    1. The problem statement, all variables and given/known data
    Does the follow serie converge:
    [tex]\sum_{n=1}^\infty sin(\frac{1}{n^2})[/tex]

    2. Relevant 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

    3. 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.
  2. jcsd
  3. Oct 8, 2007 #2


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    Science Advisor
    Homework Helper

    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.
    Last edited: Oct 8, 2007
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