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Math Help - series convegnce

  1. #1
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    series convegnce

    Hi

    How do I find limit of convegence of the series

    a>1 ; n goes from 1 to infinity

    sigma[n/(a^n)]

    Thanks
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  2. #2
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    {{\varphi }_{n}}=\frac{n}{{{a}^{n}}}\implies \sqrt[n]{{{\varphi }_{n}}}=\frac{\sqrt[n]{n}}{a}\to \frac{1}{a} as n\to\infty.

    Since \frac{1}{a}<1,\,\forall \,a>1 the series converges with given condition on a.
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  3. #3
    MHF Contributor

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    Quote Originally Posted by omer.jack View Post
    Hi

    How do I find limit of convegence of the series a>1 ; n goes from 1 to infinity sigma[n/(a^n)]

    Thanks
    to evaluate the series, differentiate 1+x+x^2 + \cdots = \frac{1}{1-x}, \ |x| < 1, to get 1 + 2x + 3x^2 + \cdots = \frac{1}{(1-x)^2}, now multiply by x and then put x=\frac{1}{a} to get \sum_{n=1}^{\infty}\frac{n}{a^n}=\frac{a}{(a-1)^2}.
    Last edited by NonCommAlg; January 12th 2009 at 02:36 PM.
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