Question: If $W$,$V$ are independent geometric random variables, what is the distribution of $Z \sim W + V$?
Please note, $W \sim Geometric(3/5)$ and $V \sim Geometric(7/12)$.
Also,
$P(W=w) = (1-\frac{3}{5})^{w}\frac{3}{5}$ for $w = 0,1,2, \dots$
$P(V=v) = (1-\frac{7}{12})^{v}\frac{7}{12}$ for $v = 0,1,2, \dots$
What I have so far: I have looked around already on mathexchange and usually the geometric pdf is different to the one I am defining (with $1-p$ to the power of $w-1$ instead of just $w$). I know to use convolution however I am unsure of how to simplify the sum, I am also unsure of if my limits are wrong.
My working:
$P(W+V=z)= \sum_{w=0}^z P(W=w)P(V=z-w) $
$= \frac{3}{5} \frac{7}{12} \sum_{w=0}^z (\frac{2}{5})^{w}(\frac{5}{12})^{z-w}$
$= \frac{3}{5} \frac{7}{12} \sum_{w=0}^z (\frac{24}{25})^{w}(\frac{5}{12})^{z}$
$= \frac{3}{5} \frac{7}{12} (\frac{5}{12})^{z} \sum_{w=0}^z (\frac{24}{25})^{w}$
My biggest question is if my limits on the sum are correct and where do I go from here?
Hint:
The sum of a geometric series is $$\sum_{i=0}^{n} a^i = \frac{a^{n+1}-1}{a-1}$$.
You can use this on the step where you're stuck.