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Thread: Dirichlet series of extended Liouville function

  1. #1
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    Dirichlet series of extended Liouville function

    I would like to prove that
    $\displaystyle \sum_{n = 1}^\infty \frac{\lambda_a}{n^s} = \frac{\zeta(s) \zeta(2s-2a)}{\zeta(s-a)} $
    where
    $\displaystyle \lambda_a(n) = \sum_{d \vert n} d^a \cdot \lambda(d) $
    and
    $\displaystyle \lambda(n) $ is Liouville's function, defined as
    $\displaystyle \lambda(1) = 1 \mbox{, and if } n=p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} \mbox{ then } \lambda(n) = (-1)^{a_1 + a_2 + \cdots + a_k} $

    Using multiplication of Dirichlet series, I think I can show that
    $\displaystyle \sum_{n = 1}^\infty \frac{\lambda (n)}{n^s} = \frac{\zeta(2s)}{\zeta(s)} $
    and that
    $\displaystyle \sum_{n = 1}^\infty \frac{\sum_{d \vert n} d^a}{n^s} = \zeta(s-a)\zeta(s) $
    but I cant see how (or if) this helps.

    Any help or advice would be appreciated.
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  2. #2
    Super Member PaulRS's Avatar
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    Well, notice that since you know : $\displaystyle \sum_{k\geq 1} {\frac{\lambda(k)}{k^s}} = \frac{\zeta(2s)}{\zeta (s)}$

    We then have: $\displaystyle \sum_{k\geq 1} {\frac{\lambda(k)}{k^{s-a}}} = \frac{\zeta(2s - 2a)}{\zeta (s-a)}$

    But here: $\displaystyle \sum_{k\geq 1} {\frac{\lambda(k)}{k^{s-a}}} = \sum_{k\geq 1} {\frac{k^a \cdot \lambda(k)}{k^s}}$ the generating function of $\displaystyle f(k) = k^a\cdot \lambda ( k)$

    Now multiplying this by $\displaystyle \zeta(s)$ we get the Dirichlet-Series-Generating-Function of $\displaystyle \sum_{d|n} {d^a\cdot \lambda( d)}$ which is what you want.
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  3. #3
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    Ahh. Thank you. I had not realized that

    $\displaystyle \sum_{k \geq 1} \frac{ \lambda (k) }{ k^{s} } = \frac{ \zeta(2s)}{ \zeta(s)} $
    would imply
    $\displaystyle \sum_{k \geq 1} \frac{ \lambda (k) }{ k^{s-a} } = \frac{ \zeta(2s-2a)}{ \zeta(s-a)} $

    Many thanks for the help.
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