Question about Digamma function and infinite sums

[MAGNIBORO]

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 don't know how It is said )
I don't know what is that true and I would like someone to show me this relation. thx

 

[thierrykauf]

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
$\psi(1-\frac{1}{k}) = - \gamma + \sum -\frac{\frac{1}{k}}{n(n-\frac{1}{k})}$
$-\frac{1}{k}\psi(1-\frac{1}{k}) = \frac{\gamma}{k} + \sum -\frac{\frac{1}{k}}{n(nk-1)}$
$-\frac{1}{k}\psi(1-\frac{1}{k}) - \frac{\gamma}{k} = \sum -\frac{\frac{1}{k}}{n(nk-1)}$

 

[MAGNIBORO]

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
$\psi(1-\frac{1}{k}) = - \gamma + \sum -\frac{\frac{1}{k}}{n(n-\frac{1}{k})}$
$-\frac{1}{k}\psi(1-\frac{1}{k}) = \frac{\gamma}{k} + \sum -\frac{\frac{1}{k}}{n(nk-1)}$
$-\frac{1}{k}\psi(1-\frac{1}{k}) - \frac{\gamma}{k} = \sum -\frac{\frac{1}{k}}{n(nk-1)}$

thnx =D