Two independent geometric random variables - proof of sum If $X$ and $Y$ are independent geometric random variables, $X \sim G(p)$ and $Y \sim G(q)$ then if $Z = X+Y$ I need to show that:

$$P[Z=z] = \frac{pq}{p-q}[(1-q)^{z-1}-(1-p)^{z-1}]$$

My attempt:
\begin{align}
P[Z=z] &= \sum_{k=1}^z P[X=k]P[Y=z-k] && \text{(not sure of my summation limits here)}  \\
&= \sum_{k=1}^z p(1-p)^{k-1} \cdot q(1-q)^{z-k-1} \\
&= \frac{pq}{(1-p)}\cdot \frac{(1-q)^z}{(1-q)}\cdot\sum_{k=1}^z (\frac{1-p}{1-q})^k \\
&= \frac{pq}{(1-p)}\cdot \frac{(1-q)^z}{(1-q)} \cdot \frac{1-(\frac{1-p}{1-q})^k}{1-\frac{1-p}{1-q}} \\
&= \frac{pq}{(1-p)}\cdot \frac{(1-q)^z}{(p-q)} \left[1 -(\frac{1-p}{1-q})^k\right] \\
&= \frac{pq}{p-q}\left[\frac{(1-q)^z}{1-p} - \frac{(1-p)^z}{1-p}\right]
\end{align}
The result is pretty close to the answer. The only discrepancy is the first expression in the parentheses, $\frac{(1-q)^z}{1-p}$ which should be $\frac{(1-q)^z}{1-q}$.  
Could someone please have a look at my working and show me where I have gone wrong? Or perhaps a neater calculation? Thanks!  
PS
I did check similar questions answered here, but couldn't find anything relevant to my problem.
Is the sum of two independent geometric random variables with the same success probability a geometric random variable?
How to compute the sum of random variables of geometric distribution
 A: I think you made a mistake in the calculation. First geometric random variable is at least 1, hence $z\ge 2$.
\begin{align*}
P[Z=z] &= \sum_{k=1}^{z-1} P[X=k]P[Y=z-k]\\
&=\sum_{k=1}^{z-1} p(1-p)^{k-1} \cdot q(1-q)^{z-k-1}\\  
&=\frac{pq}{(1-p)} (1-q)^{z-1}\sum_{k=1}^{z-1} (\frac{1-p}{1-q})^k  \\
&=\frac{pq}{(1-p)} (1-q)^{z-1} \cdot \frac{1-p}{1-q}\frac{1-(\frac{1-p}{1-q})^{z-1}}{1-\frac{1-p}{1-q}}\\  
&=\frac{pq}{p-q} (1-q)^{z-1} \cdot [1 -(\frac{1-p}{1-q})^{z-1}]\\
&=\frac{pq}{p-q}[(1-q)^{z-1} - (1-p)^{z-1}],
\end{align*}
which is the desired result.
A: I like using the probability generating function for this:  you know the pgf for the sum is the product of the pgfs for the two random variables. So

$$p_X(t)p_Y(t)={pqt^2\over (1-t(1-p))(1-t(1-q))}=$$

Applying partial fractions to just the reciprocal terms and setting consecutively $t=(1-p)^{-1}$ then $t=(1-q)^{-1}$ gives
$$A(1-t(1-q))+B(1-t(1-p))=1\iff A={1-p\over q-p}, B = -{1-q\over q-p}$$
leaving us with

$$p_{X+Y}(t) = {pqt^2\over q-p}\left({(1-p)\over 1-t(1-p)}-{1-q\over 1-t(1-q)}\right)$$
  $$=\sum_{k=0}^\infty {pq\over q-p}((1-p)^{k+1}-(1-q)^{k+1})t^{k+2}$$

and reindexing appropriately gives

$$p_{X+Y}(t) = \sum_{i=2}^\infty {pq\over p-q}\Big((1-q)^{k-1}-(1-p)^{k-1}\Big)t^k$$

as desired.
