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I Question about Digamma function and infinite sums

  1. Sep 23, 2016 #1
    hi, I'm solving solving a problem about sums of zeta function and I'm come to the following conclusion
    $$\sum _{n=2}^{\infty }{\frac {\zeta \left( n \right) }{{k}^{n}}}=
    \sum _{s=1}^{\infty } \left( {\it ks} \left( {\it ks}-1 \right)
    \right) ^{-1}=\int_{0}^{1}\!{\frac {{u}^{k-2}}{\sum _{i=0}^{k-1}{u}^{i}}}\,{\rm d}u,\: \:\: \: \forall k, \: \:k\geq 2 \: \: and \: \: k\in \mathbb{N}$$

    and from wolfram alpha tell me that

    $$=-{\frac {1}{k} \left( \psi \left( 1-{k}^{-1} \right) +\gamma \right) }$$
    where the ##\psi## is the "zero" digamma function (I dont know how It is said )
    I dont know what is that true and I would like someone to show me this relation. thx
     
  2. jcsd
  3. Sep 26, 2016 #2

    thierrykauf

    User Avatar
    Gold Member

    The Digamma function is defined as ##\psi(z) = \frac{\Gamma'(z)}{\Gamma(z)}## and has many expressions as sums, one of which is ##\psi(z+1) = -\gamma + \sum \frac{z}{n(n + z)}##
    Replace z by -1/k you get
    ##\psi(1-\frac{1}{k}) = - \gamma + \sum -\frac{\frac{1}{k}}{n(n-\frac{1}{k})}##
    ##-\frac{1}{k}\psi(1-\frac{1}{k}) = \frac{\gamma}{k} + \sum -\frac{\frac{1}{k}}{n(nk-1)}##
    ##-\frac{1}{k}\psi(1-\frac{1}{k}) - \frac{\gamma}{k} = \sum -\frac{\frac{1}{k}}{n(nk-1)}##
     
  4. Sep 30, 2016 #3
    thnx =D
     
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