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Math Help - Convergence of a series

  1. #1
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    Convergence of a series

    Hi everyone.
    I need to prove that if 0 < a < 1 , then the series \sum_{n=1}^{\infty} (1-\frac{1}{n^a})^n converges.
    I tried to use the limit test and the comparison test but I couldn't manage to.
    Can someone give me a hint how to prove that?

    Thanks in advance!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Gok2 View Post
    Hi everyone.
    I need to prove that if 0 < a < 1 , then the series \sum_{n=1}^{\infty} (1-\frac{1}{n^a})^n converges.
    I tried to use the limit test and the comparison test but I couldn't manage to.
    Can someone give me a hint how to prove that?

    Thanks in advance!
    Notice that \left(1-\frac{1}{n^\alpha}\right)^n=e^{n\ln\left(1-\frac{1}{n^\alpha}\right)}. But, notice that since 0<\frac{1}{n}\leqslant 1 we may apply the series to the natural log to obtain that e^{n\ln\left(1-\frac{1}{n^\alpha}\right)}=e^{-n\left(\frac{1}{n^\alpha}+\frac{1}{2n^{2\alpha}}+\  cdots\right)}\approx e^{-n^{1-\alpha}} so what happens if 0<\alpha<1?
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