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Math Help - Exchangeable of the limitation

  1. #1
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    Question Exchangeable of the limitation

    \lim_{x\rightarrow1-0}\sum_{n=1}^{\infty}\frac{x^{n-1}}{nln(n+1)} = \sum_{n=1}^{\infty}\lim_{x\rightarrow1-0}\frac{x^{n-1}}{nln(n+1)}

    Is the above equation right? in the other words, can we take the limitation into the summation?
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  2. #2
    Super Member Rebesques's Avatar
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    Yes, as long as the quantity considered is positive. The worse thing that could happen is that both iterated limits are infinite.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Xingyuan View Post
    \lim_{x\rightarrow1-0}\sum_{n=1}^{\infty}\frac{x^{n-1}}{nln(n+1)} = \sum_{n=1}^{\infty}\lim_{x\rightarrow1-0}\frac{x^{n-1}}{nln(n+1)}

    Is the above equation right? in the other words, can we take the limitation into the summation?
    This would be true if \displaystyle \sum_{n=1}^{\infty}\frac{1}{n\log(n+1)} converged. As it is stated it doesn't make much sense. But, for example \displaystyle \lim_{x\to 1^{-1}}\sum_{n=1}^{\infty}\frac{x^{n-1}}{n^2\log(n+1)}=\sum_{n=1}^{\infty}\lim_{x\to 1^{-}}\frac{x^{n-1}}{n^2\log(n+1)}. This is because \displaystyle \frac{x^{n-1}}{n^2\log(n+1)} has radius of convergence 1 from where you can apply Abel's Limit Theorem.
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