Your ultimate goal is to prove that
$\int\limits_0^1 {{t^{1/\alpha }}\int\limits_0^1 {\frac{{vz}}{{1 - tvz}}dv} dt} = \frac{1}{\alpha }\sum\limits_{n = 2}^\infty {\frac{{{z^{n - 1}}}}{{n\left( {\alpha \left( {n - 1} \right) + 1} \right)}}} $
This can be done by geometric series since we're integrating over $(0,1$).
$\frac{{vz}}{{1 - tvz}} = vz\sum\limits_{k = 0}^\infty {{t^k}{v^k}{z^k}} = \sum\limits_{k = 0}^\infty {{t^k}{v^{k + 1}}{z^{k + 1}}} $
Thus we have
$\int\limits_0^1 {\frac{{vz}}{{1 - tvz}}} dv = \sum\limits_{k = 0}^\infty {\frac{{{t^k}{z^{k + 1}}}}{{k + 2}}} $
Moving on we get:
$\int\limits_0^1 {\sum\limits_{k = 0}^\infty {\frac{{{t^{k + 1/\alpha }}{z^{k + 1}}}}{{k + 2}}} } dt = \sum\limits_{k = 0}^\infty {\frac{{{z^{k + 1}}}}{{\left( {k + 2} \right)\left( {k + 1 + 1/\alpha } \right)}}} $
Rearranging to $k=2$ we get
$\eqalign{ & \sum\limits_{k = 2}^\infty {\frac{{{z^{k - 2 + 1}}}}{{\left( {k - 2 + 2} \right)\left( {k - 2 + 1 + 1/\alpha } \right)}}} \cr & \sum\limits_{k = 2}^\infty {\frac{{{z^{k - 1}}}}{{k\left( {k - 1 + 1/\alpha } \right)}}} \cr & \frac{1}{\alpha }\sum\limits_{k = 2}^\infty {\frac{{{z^{k - 1}}}}{{k\left( {\alpha \left( {k - 1} \right) + 1} \right)}}} \cr} $
which is what you wanted.