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Math Help - Convergence of Infinite Series

  1. #1
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    Convergence of Infinite Series

    Hi there,

    I'm trying to prove that if

    An <= (1 - 1/n^a)^n

    (when 0 < a < 1)

    Then the series ∑An converges.

    I managed to get to the form of

    ∑ (1/e)^(n^b)

    (where 0<b<1)

    But I can't even prove that this more simple series converges.

    Could someone help me please?

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by Prism View Post
    I managed to get to the form of

    ∑ (1/e)^(n^b)

    (where 0<b<1)

    But I can't even prove that this more simple series converges.
    For instance, you can say that \frac{n^2}{e^{n^b}}=\frac{(n^b)^{2/b}}{e^{n^b}}\to 0 because \frac{x^{2/b}}{e^x}\to_{x\to\infty} 0. Therefore, for large n, we have \frac{n^2}{e^{n^b}}\leq 1, thus 0<\frac{1}{e^{n^b}}\leq \frac{1}{n^2} hence the convergence.
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  3. #3
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    Thank you very much!
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  4. #4
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    I assume the original problem is:
    If A_n \leq (1 - \frac{1}{n^a})^n where 0 < a < 1 then ∑An converges.

    Just another approach:

    Prove that \sum_{n=1}^{\infty} (1 - \frac{1}{n^a})^n is convergent then by the comparison test also ∑An converges.

    or it is the same approach?
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by p0oint View Post
    I assume the original problem is:
    If A_n \leq (1 - \frac{1}{n^a})^n where 0 < a < 1 then ∑An converges.

    Just another approach:

    Prove that \sum_{n=1}^{\infty} (1 - \frac{1}{n^a})^n is convergent then by the comparison test also ∑An converges.

    or it is the same approach?
    I would note that \left(1-\frac{1}{n^\alpha}\right)^n=e^{n\ln\left(1-\frac{1}{n^\alpha}\right)}. Then, for sufficiently large n we have that n^\alpha>1\implies \frac{1}{n^\alpha}<1 and thus \ln\left(1-\frac{1}{n^\alpha}\right)=-\sum_{m=1}^{\infty}\frac{1}{n^{\alpha m}m} which clearly behaves as \frac{-1}{n^\alpha} for large n and thus \left(1-\frac{1}{n^\alpha}\right)\overset{n\to\infty}{\sim  }e^{-n^{1-\alpha}}....which I just realized is what you did and Laurent finished. My apologies.
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