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Math Help - convergent/divergent series

  1. #1
    Member disclaimer's Avatar
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    convergent/divergent series

    Hi,

    How to check whether \sum_{n=1}^{\infty}{n\sin}\frac{1}{n^3} is convergent/divergent? I could show that \sum_{n=1}^{\infty}{n\sin}\frac{1}{n^2} is divergent, but I don't know how to start with the first one... Won't someone drop a hint, please?

    Input appreciated.
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  2. #2
    Super Member PaulRS's Avatar
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    Since \sin(x)\approx{x} when x\rightarrow{0}

    we have n\cdot{\sin\left(\frac{1}{n^3}\right)}\approx{\fra  c{1}{n^2}} when n\rightarrow{+\infty} ( \lim_{n\rightarrow{+\infty}}\frac{n\cdot{\sin\left  (\frac{1}{n^3}\right)}}{\frac{1}{n^2}}=1 )

    But: \sum_{n=1}^{\infty}{\frac{1}{n^2}} does converge, so, by the Limit Comparison Test, \sum_{n=1}^{\infty}{n\cdot{\sin\left(\frac{1}{n^3}  \right)}} converges

    Again: \lim_{n\rightarrow{+\infty}}\frac{n\cdot{\sin\left  (\frac{1}{n^2}\right)}}{\frac{1}{n}}=1

    and \sum_{n=1}^{\infty}{\frac{1}{n}} diverges, thus: \sum_{n=1}^{\infty}{n\cdot{\sin\left(\frac{1}{n^2}  \right)}} diverges
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  3. #3
    Member disclaimer's Avatar
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    Great! The way you showed seems to be valid for any positive \alpha.

    <br />
\sum_{n=1}^{\infty}{n\sin}\frac{1}{n^\alpha}<br />
    Last edited by disclaimer; March 8th 2008 at 01:37 AM.
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