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Thread: convergent/divergent series

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

    Hi,

    How to check whether $\displaystyle \sum_{n=1}^{\infty}{n\sin}\frac{1}{n^3}$ is convergent/divergent? I could show that $\displaystyle \sum_{n=1}^{\infty}{n\sin}\frac{1}{n^2}$ 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 $\displaystyle \sin(x)\approx{x}$ when $\displaystyle x\rightarrow{0}$

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

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

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

    and $\displaystyle \sum_{n=1}^{\infty}{\frac{1}{n}}$ diverges, thus: $\displaystyle \sum_{n=1}^{\infty}{n\cdot{\sin\left(\frac{1}{n^2} \right)}}$ 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 $\displaystyle \alpha$.

    $\displaystyle
    \sum_{n=1}^{\infty}{n\sin}\frac{1}{n^\alpha}
    $
    Last edited by disclaimer; Mar 8th 2008 at 01:37 AM.
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