Trying to understand to equal sums. Why is it that? 
$$z + \sum_{2 \le k \le n}^{\infty} \frac{q^{n-1}}{n-1} k(k-1)p^{n-k} \dbinom{2n-k-2}{n-2}z^n = \sum_{j, k \ge 0}^{\infty} \frac{q^{j+k-1}}{j+k-1} k(k-1)p^j \dbinom{2j+k-2}{j}z^{j+k}. $$
I get that you can just plug $j+k=n$, but where does the $z$ appears in the latter?
 A: 
We start with the right-hand side and obtain
  \begin{align*}
\color{blue}{\sum_{j,k\geq 0}}&\color{blue}{\frac{q^{j+k-1}}{j+k-1}k(k-1)p^j\binom{2j+k-2}{j}z^{j+k}}\\
&=\sum_{n=0}^\infty\left(\sum_{{j+k=n}\atop{j,k\geq 0}}\frac{q^{j+k-1}}{j+k-1}k(k-1)p^j\binom{2j+k-2}{j}\right)z^n\tag{1}\\
&=\sum_{n=0}^\infty\sum_{k=0}^n\frac{q^{n-1}}{n-1}k(k-1)p^{n-k}\binom{2n-k-2}{n-k}z^n\tag{2}\\
&=\sum_{0\leq k\leq n\leq \infty}\frac{q^{n-1}}{n-1}k(k-1)p^{n-k}\binom{2n-k-2}{n-2}z^n\tag{3}\\
&\,\,\color{blue}{=z+\sum_{2\leq k\leq n\leq \infty}\frac{q^{n-1}}{n-1}k(k-1)p^{n-k}\binom{2n-k-2}{n-2}z^n}\tag{4}\\
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we reorder the summands by introducing $n=j+k, n\geq 0$. 

*In (2) we eliminate the index $j$ by substituting $j=n-k$.

*In (3) we rewrite the index region and use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

*In (4) we observe that the summands with indices $(n,k)\in\{(0,0),(0,1),(1,0)\}$ vanish due to the factor $k(k-1)$, whereas in the case $(n,k)=(1,1)$ the expression $\frac{k-1}{n-1}$ cancels, leaving $z$. 
