X and Y are independent and geometrically distributed random variables with $$ P(X = m) = p(1-p)^{m}, m=0,1,2... $$ $$ P(Y = n) = p(1-p)^{n}, n=0,1,2... $$ To find the probability mass function (PMF) of Z= X+Y, I should find the moment-generating function of Z $$ M_{Z} = M_{X}M_{Y} = \left (\frac{p}{1-(1-p)e^{t}}\right)^2 $$
My question is: How to find the PMF of $M_{Z}$?
We have $X+Y=z$ if $X=0$ and $Y=z$, or $X=1$ and $Y=z-1$, and so on up to $X=z$ and $Y=0$. Thus $$\Pr(Z=z)=\sum_{k=0}^z p(1-p)^k p(1-p)^{z-k}.$$ This simplifies to $$\sum_{k=0}^z p^2(1-p)^z.$$ This simplifies to $(z+1)p^2(1-p)^z$.
Remark: We can use the moment-generating function approach. However, once we have found the mgf, we need to identify the distribution. That is easy only if we have a "dictionary" of moment-generating functions.
A sum of independent geometric random variables with the same "$p$" has negative binomial distribution. It is the number of "failures" until the $n$-th success. The probability that this sum is equal to $z$ is equal to $$\binom{z+1}{n-1}p^{n-1}(1-p)^{(z+1)-(n-1)}p.$$ For we must have exactly $n-1$ successes in the first $z+1$ trials, and then a success. Ours is the case $n=2$.