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Math Help - Infinite series (8)

  1. #1
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    Infinite series (8)

    Evaluate \sum_{n=1}^{\infty} (-1)^n \frac{\ln n}{n}.
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  2. #2
    Super Member PaulRS's Avatar
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    Consider: \sum_{k=1}^{2n}{(-1)^k\cdot \tfrac{\log(k)}{k}}=2\sum_{k=1}^{n}{\tfrac{\log(2k  )}{2k}}-\sum_{k=1}^{2n}{\tfrac{\log(k)}{k}}

    Note that: \sum_{k=1}^{n}{\tfrac{\log(2k)}{2k}}=\frac{1}{2}\c  dot \left(\log(2)\cdot \text{H}_n+\sum_{k=1}^{n}{\tfrac{\log(k)}{k}}\righ  t)

    So we have: \sum_{k=1}^{2n}{(-1)^k\cdot \tfrac{\log(k)}{k}}= \left(\log(2)\cdot \text{H}_n+\sum_{k=1}^{n}{\tfrac{\log(k)}{k}}\righ  t)-\sum_{k=1}^{2n}{\tfrac{\log(k)}{k}} (1)

    Now: \sum_{k=1}^{n}{\tfrac{\log(k+1)}{k+1}}=\int_0^n\fr  ac{\log(x+1)}{x+1}dx+\int_0^n\left\{ x \right\}\cdot \left(\frac{1-\log(x+1)}{(x+1)^2}\right)dx (see here )

    The integral on the right clearly converges, thus: \sum_{k=1}^{n}{\tfrac{\log(k+1)}{k+1}}=\frac{\log^  2(n+1)}{2}+C+o(1) where C=\int_0^{+\infty}\left\{ x \right\}\cdot \left(\frac{1-\log(x+1)}{(x+1)^2}\right)dx . Hence \sum_{k=1}^{n}{\tfrac{\log(k)}{k}}=\frac{\log^2(n)  }{2}+C+1+o(1)

    Recall that: \text{H}_n=\log(n)+\gamma+o(1).

    Thus, going back to (1) : \sum_{k=1}^{2n}{(-1)^k\cdot \tfrac{\log(k)}{k}}=\log(2)\cdot (\log(n)+\gamma+o(1))+ \left(\frac{\log^2(n)}{2}+C+1+o(1)\right)-\left(\frac{\log^2(2n)}{2}+C+1+o(1)\right)

    \sum_{k=1}^{2n}{(-1)^k\cdot \tfrac{\log(k)}{k}}=\log(2)\cdot (\log(n)+\gamma)+ \left(\frac{\log^2(n)}{2}\right)-\left(\frac{\log^2(2)+2\log(2)\log(n)+\log^2(n)}{2  }\right)+o(1)

    So finally: \sum_{k=1}^{2n}{(-1)^k\cdot \tfrac{\log(k)}{k}}=\log(2)\cdot \gamma-\tfrac{1}{2}\cdot \log^2(2)+o(1) Hence your series is: \sum_{k=1}^{\infty}{(-1)^k\cdot \tfrac{\log(k)}{k}}=\log(2)\cdot \gamma-\tfrac{1}{2}\cdot \log^2(2)
    Last edited by PaulRS; December 8th 2009 at 06:08 AM. Reason: Fixing a typo
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by PaulRS View Post
    Consider: \sum_{k=1}^{2n}{(-1)^k\cdot \tfrac{\log(k)}{k}}=2\sum_{k=1}^{n}{\tfrac{\log(2k  )}{2k}}-\sum_{k=1}^{2n}{\tfrac{\log(k)}{k}}

    Note that: \sum_{k=1}^{n}{\tfrac{\log(2k)}{2k}}=\frac{1}{2}\c  dot \left(\log(2)\cdot \text{H}_n+\sum_{k=1}^{n}{\tfrac{\log(k)}{k}}\righ  t)

    So we have: \sum_{k=1}^{2n}{(-1)^k\cdot \tfrac{\log(k)}{k}}= \left(\log(2)\cdot \text{H}_n+\sum_{k=1}^{n}{\tfrac{\log(k)}{k}}\righ  t)-\sum_{k=1}^{2n}{\tfrac{\log(k)}{k}} (1)

    Now: \sum_{k=1}^{n}{\tfrac{\log(k+1)}{k+1}}=\int_0^n\fr  ac{\log(x+1)}{x+1}dx+\int_0^n\left\{ x \right\}\cdot \left(\frac{1-\log(x+1)}{(x+1)^2}\right)dx (see here )

    The integral on the right clearly converges, thus: \sum_{k=1}^{n}{\tfrac{\log(k+1)}{k+1}}=\frac{\log^  2(n+1)}{2}+C+o(1) where C=\int_0^{+\infty}\left\{ x \right\}\cdot \left(\frac{1-\log(x+1)}{(x+1)^2}\right)dx . Hence \sum_{k=1}^{n}{\tfrac{\log(k)}{k}}=\frac{\log^2(n)  }{2}+C+1+o(1)

    Recall that: \text{H}_n=\log(n)+\gamma+o(1).

    Thus, going back to (1) : \sum_{k=1}^{2n}{(-1)^k\cdot \tfrac{\log(k)}{k}}=\log(2)\cdot (\log(n)+\gamma+o(1))+ \left(\frac{\log^2(n)}{2}+C+1+o(1)\right)-\left(\frac{\log^2(2n)}{2}+C+1+o(1)\right)

    \sum_{k=1}^{2n}{(-1)^k\cdot \tfrac{\log(k)}{k}}=\log(2)\cdot (\log(n)+\gamma)+ \left(\frac{\log^2(n)}{2}\right)-\left(\frac{\log^2(2)+2\log(2)\log(n)+\log^2(n)}{2  }\right)+o(1)

    So finally: \sum_{k=1}^{2n}{(-1)^k\cdot \tfrac{\log(k)}{k}}=\log(2)\cdot \gamma-\tfrac{1}{2}\cdot \log^2(2)+o(1) Hence your series is: \sum_{k=1}^{\infty}{(-1)^k\cdot \tfrac{\log(k)}{k}}=\log(2)\cdot \gamma-\tfrac{1}{2}\cdot \log^2(2)
    This is great!

    PaulRS, I'd like to pick your brain!
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