Proving the equivalence of a sum and a double integral Based on "Certain Subclass of Starlike Functions" journal by Chun-Yi and Shi-Qiong Zhou in 2007 (Science Direct), I found difficulties to understand the proof in Theorem 3 where they have verified: 
$$ 1+2(1-\beta)  \displaystyle\sum\limits_{n=2}^\infty  \frac{z^{n-1}}{n(\alpha(n-1)+1)} = 1 +\frac{2(1-\beta)}{\alpha} \int_0^1 \! t^{\frac{1}{\alpha}} \,\int_0^1 \! \frac{vz}{1-tvz} \, \mathrm{d} v \ \mathrm{d} t$$ 
Could someone give me the idea of how to prove it?
Thank you.
 A: 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. 
A: Subtract $1$, then multiply by $\frac{\alpha}{2(1-\beta)}$ on both sides. Expand (using a sum from $n=2$ to $\infty$ as in the desired result):
$$
\frac{vz}{1-tvz}=\sum_{n=2}^{\infty}t^{n-2}(vz)^{n-1}
$$
Exchange summation and inside integration (inside the radius of convergence, where the series is absolutely convergent), and use the fact that $\int_0^1v^kdv=\frac{1}{k+1}$ (for $k>-1$), then add the exponents of $t$, swap summation and integration again and use the same fact.
