Trying to understand to equal sums.

Question

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?

Answer 1

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 $z$.