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Math Help - Summation of a series giving logarithm

  1. #1
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    Summation of a series giving logarithm

    I need to prove that \sum_{i=1}^{\infty}\sum_{j=1}^{i-1}\frac{(-1)^i}{i j}=\frac{1}{2}\ln ^2 2. The obvious way would be to compare with the series on the r.h.s. which is \frac{1}{2}\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\  frac{(-1)^{(i+j)}}{i j}, but I just can't show that they're the same. Any suggestions?
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  2. #2
    Junior Member Renji Rodrigo's Avatar
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     \sum\limits^{\infty}_{k=1}\sum\limits^{k-1}_{j=1}\frac{(-1)^{k}}{(k)j}.


    We have
     \sum\limits^{\infty}_{k=1}\sum\limits^{k-1}_{j=1}\frac{(-1)^{k}}{(k)j}=\sum\limits^{\infty}_{k=0}\sum\limit  s^{k}_{j=1}\frac{(-1)^{k+1}}{(k+1)j}.

    when k=0in the second series we have  \sum\limits^{k}_{j=1}\frac{(-1)^{k+1}}{(k+1)j}=0 because the sum is empty, so we can write

    \sum\limits^{\infty}_{k=1}\sum\limits^{k}_{j=1}\fr  ac{(-1)^{k+1}}{(k+1)j}
    reverting the order of summation

     \sum\limits^{\infty}_{k=1}\sum\limits^{k}_{j=1}\fr  ac{(-1)^{k+1}}{(k+1)j}=  \sum\limits^{\infty}_{j=1}\sum\limits^{\infty}_{k=  j}\frac{(-1)^{k+1}}{(k+1)j}=


    =\sum\limits^{\infty}_{j=1}\sum\limits^{\infty}_{k  =0}\frac{(-1)^{k+1+j}}{(k+1+j)j}=


    = \sum\limits^{\infty}_{j=1}\frac{(-1)^{j+1}}{j}\sum\limits^{\infty}_{k=0}(-1)^k \int^{1}_{0}x^{k+j}dx=

    =\int^{1}_{0}  \sum\limits^{\infty}_{j=1}\frac{(-1)^{j+1}x^j}{j}\sum\limits^{\infty}_{k=0}  (-x)^{k}dx=

     \int^{1}_{0}\frac{ln(x+1)}{x+1}=\frac{\ln^2 (2)}{2}.
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  3. #3
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    Thank you very much! Your method of dealing with sums has enlightened me.
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  4. #4
    Junior Member Renji Rodrigo's Avatar
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    you're welcome!
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