Let $X_n$ be the size of the $n$-th generation of a branching process, with family size probability generating function $G(s)$. Let $X_0 = 1$.
Suppose the family-mass size mass function is $P(X_1 = k) = 2^{-k-1}$ for $k=0,1,2,\dots$. What is the moment generating function of $X_n/n$? Hence, or otherwise, show that for $x\geq 0$, $$P(X_n/n \geq x \mid X_n > 0)\rightarrow e^{-x}\quad\text{ as }n\rightarrow\infty.$$
I have established that $G(s) = \frac{1}{2-s}$ and that the PGF of $X_n$ is given by $$G_n(s) = \frac{n-(n-1)s}{n+1-ns}$$ by a simple induction. I also know that $P(X_n = 0) = G_n(0) = \frac{n}{n+1}$.
Am I right to say that the MGF of $X_n/n$ is $$M_{X_n/n}(t) = \mathbb{E}(e^{\frac{X_n}{n}t}) = \mathbb{E}((e^{\frac{t}{n}})^{X_n}) = G_n(e^{t/n}) = \frac{n-(n-1)e^{t/n}}{n+1-ne^{t/n}}?$$ I also tried to use the continuity theorem to see what $M_{X_n/n}(t)$ tends to as $n\rightarrow\infty$ but I got that the MGF tends to 1 which is basically the identically zero variable. How should I go about establishing the result on the probability?