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Math Help - On the Basel problem

  1. #1
    Senior Member bkarpuz's Avatar
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    Exclamation On the Basel problem

    Dear MHF members,

    the problem is related to the so-called Basel sum.

    Let
    S:=1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots.
    Show that
    1-\frac{1}{2^{2}}-\frac{1}{4^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}-\frac{1}{8^{2}}-\frac{1}{10^{2}}+\frac{1}{11^{2}}+\frac{1}{13^{2}}-\cdot-\cdot+\cdot+\cdots=\frac{4}{9}S.

    Thanks.
    bkarpuz
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  2. #2
    Member sbhatnagar's Avatar
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    Re: On the Basel problem

    Given that : \sum_{n=1}^{\infty}\frac{1}{n^2}=S=\frac{\pi^2}{6}

    \begin{align*} k &=1-\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{7^2}-\cdots \\ &= \left( 1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2} +\cdots \right) - \left( \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2} +\cdots \right)-\left(\frac{1}{3^2}-\frac{1}{6^2}+\frac{1}{9^2} -\cdots\right)\\ \end{align*}

    Note that :

    [1] \quad \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2} +\cdots=\sum_{n=1}^{\infty}\frac{1}{(2n)^2}=\frac{  1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{S}{4}

    [2] \quad 1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2} +\cdots=S-\left( \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2} +\cdots \right)= S-\frac{S}{4}=\frac{3S}{4}

     \begin{align*}[3] \quad \frac{1}{3^2}-\frac{1}{6^2}+\frac{1}{9^2} -\cdots &= \left( \frac{1}{3^2}+\frac{1}{9^2}+\frac{1}{15^2}+\cdots \right)-\left( \frac{1}{6^2}+\frac{1}{12^2}+\frac{1}{18^2}+\cdots \right) \\ &=\left( \sum_{n=1}^{\infty}\frac{1}{(3n)^2}-\sum_{n=1}^{\infty}\frac{1}{(6n)^2} \right) - \sum_{n=1}^{\infty}\frac{1}{(6n)^2} \\ &= \frac{S}{9}-\frac{S}{18} \\ &= \frac{S}{18}\end{align*}

    Hence :

    \begin{align*} k &= \left( 1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2} +\cdots \right) - \left( \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2} +\cdots \right)-\left(\frac{1}{3^2}-\frac{1}{6^2}+\frac{1}{9^2} -\cdots\right) \\ &=\frac{3S}{4}-\frac{S}{4}-\left(\frac{1}{3^2}-\frac{1}{6^2}+\frac{1}{9^2} -\cdots\right) \\ &= \frac{S}{2}-\left(\frac{1}{3^2}-\frac{1}{6^2}+\frac{1}{9^2} -\cdots\right) \\ &=\frac{S}{2}-\frac{S}{18} \\&= \frac{4S}{9}\end{align*}


    Happy New Year!
    sbhatnagar
    Last edited by sbhatnagar; December 29th 2011 at 01:58 AM.
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