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Thread: Infinite series (8)

  1. #1
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    Infinite series (8)

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

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

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

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

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

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

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

    $\displaystyle \sum_{k=1}^{2n}{(-1)^k\cdot \tfrac{\log(k)}{k}}=\log(2)\cdot (\log(n)+\gamma)+$$\displaystyle \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: $\displaystyle \sum_{k=1}^{2n}{(-1)^k\cdot \tfrac{\log(k)}{k}}=\log(2)\cdot \gamma-\tfrac{1}{2}\cdot \log^2(2)+o(1)$ Hence your series is: $\displaystyle \sum_{k=1}^{\infty}{(-1)^k\cdot \tfrac{\log(k)}{k}}=\log(2)\cdot \gamma-\tfrac{1}{2}\cdot \log^2(2)$
    Last edited by PaulRS; Dec 8th 2009 at 05:08 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: $\displaystyle \sum_{k=1}^{2n}{(-1)^k\cdot \tfrac{\log(k)}{k}}=2\sum_{k=1}^{n}{\tfrac{\log(2k )}{2k}}-\sum_{k=1}^{2n}{\tfrac{\log(k)}{k}}$

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

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

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

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

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

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

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

    PaulRS, I'd like to pick your brain!
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