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Math Help - Riemann Zeta Identity

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Riemann Zeta Identity

    For  \Re(s)>1 show  \sum_{n=1}^\infty \frac{\tau\left(n^2\right)}{n^s} = \frac{\zeta^3(s)}{\zeta(2s)} where  \tau(m) is the number of divisors of  m .
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  2. #2
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    If  m = p_1^{a_1}p_2^{a_2}p_2^{a_3}... then  \tau(m) = (1 + a_1)(1 + a_2 )( 1 + a_3) ...


    and  \tau(m^2) = (1 + 2a_1 )( 1 + 2a_2 )( 1 + 2a_3) ...


    and the sum  \sum_{n=1}^{\infty} \frac{\tau(n^2)}{n^s}

     = \prod_{p_i \in \mathbb{P} }\left( \sum_{k=0}^{\infty} \frac{2k+1}{p_i^{ks}} \right)

    Since

     \frac{1}{1-x} = 1 + x + x^2 + ....

    By taking derivative ,

     \frac{x}{(1-x)^2} = x + 2x^2 + 3x^3 + ...

    Then  \sum_{k=0}^{\infty} (2k+1)x^k = \frac{2x}{(1-x)^2} + \frac{1}{1-x} = \frac{2x+1-x}{(1-x)^2} = \frac{1 + x}{(1-x)^2}

    Let  x = \frac{1}{p_i^s }

    we have  \sum_{k=0}^{\infty} \frac{2k+1}{p_i^{ks}} = \frac{1 + 1/p_i^s }{ \left( 1 - \frac{1}{p_i^s} \right)^2 }

     = \frac{1 - 1/p_i^{2s} }{ \left( 1 - \frac{1}{p_i^s} \right)^3 }


    Therefore ,

    \sum_{n=1}^{\infty} \frac{\tau(n^2)}{n^s} = \prod_{p_i \in \mathbb{P} } \frac{1 - 1/p_i^{2s} }{ \left( 1 - \frac{1}{p_i^s} \right)^3 }  = \frac{ \zeta^3(s)}{\zeta(2s)} since

    \frac{1}{ \zeta(z)} = \prod_{p_i \in \mathbb{P} } (1- \frac{1}{p_i^z} )

    *  \mathbb{P} denotes the set of all prime numbers .
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