To find the probability $P(X + Y = Z)$, we can sum over all possible values of $X$ and $Y$. Since $X$, $Y$, and $Z$ are independent random variables, the joint probability $P(X=x, Y=y, Z=x+y)$ is the product of their individual marginal probabilities.
The problem states that $X, Y \geq 1$. Therefore, we need to evaluate the following double sum:
$$ P(X + Y = Z) = \sum_{x=1}^{\infty} \sum_{y=1}^{\infty} P(X=x) \cdot P(Y=y) \cdot P(Z=x+y) $$
$$ \Rightarrow P(X+Y=Z) = \sum_{x=1}^{\infty} \sum_{y=1}^{\infty} \left[ \frac{1}{2}\left(\frac{1}{2}\right)^{x-1} \cdot \frac{1}{3}\left(\frac{2}{3}\right)^{y-1} \cdot \frac{1}{4}\left(\frac{3}{4}\right)^{x+y-1} \right] $$
$$ \Rightarrow P(X+Y=Z) = \sum_{x=1}^{\infty} \sum_{y=1}^{\infty} \left[ \frac{1}{24} \cdot \frac{3}{4} \cdot \left(\frac{1}{2}\right)^{x-1} \left(\frac{3}{4}\right)^{x-1} \cdot \left(\frac{2}{3}\right)^{y-1} \left(\frac{3}{4}\right)^{y-1} \right] $$
$$ \Rightarrow P(X+Y=Z) = \frac{3}{96} \sum_{x=1}^{\infty} \sum_{y=1}^{\infty} \left[ \left(\frac{1}{2} \cdot \frac{3}{4}\right)^{x-1} \cdot \left(\frac{2}{3} \cdot \frac{3}{4}\right)^{y-1} \right] $$
$$ \Rightarrow P(X+Y=Z) = \frac{1}{32} \sum_{x=1}^{\infty} \sum_{y=1}^{\infty} \left[ \left(\frac{3}{8}\right)^{x-1} \cdot \left(\frac{1}{2}\right)^{y-1} \right] $$
$$ \Rightarrow P(X+Y=Z) = \frac{1}{32} \left( \sum_{x=1}^{\infty} \left(\frac{3}{8}\right)^{x-1} \right) \left( \sum_{y=1}^{\infty} \left(\frac{1}{2}\right)^{y-1} \right) $$
Now, we know that for an infinite geometric series $\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}$ (which is equivalent to our sums since they start at an exponent of $0$ when $x,y=1$):
$$ \Rightarrow P(X+Y=Z) = \frac{1}{32} \left( \frac{1}{1 - \frac{3}{8}} \right) \left( \frac{1}{1 - \frac{1}{2}} \right) $$
$$ \Rightarrow P(X+Y=Z) = \frac{1}{32} \cdot \left( \frac{8}{5} \right) \cdot (2) = \frac{16}{160} $$
$$ \boxed {\therefore P(X+Y=Z) = \frac{1}{10}} $$
