Convergence in distribution of Geometric r.v.s sequence 
Let $X_n$ be a sequence of r.v.s with Geometric distribution with parameter $p_n=\frac{5}{n}$. Prove that $Z_n = \frac{X_n}{n}$ converges in distribution. What is its limiting distribution?


I'm studying for my math exam(from all three previous years) but I completely forgot how would one do such task. I guess we'd like to use CLT here but no further progress has been made.
Thanks for any help.
 A: There is no role of CLT here. CLT gives the convergence to the normal distribution, and it is clear here that the limiting distribution must be supported only on the positive half of the real line, it can't be normal.
Since the geometric random variables are easy to analyze, that is, the probability of simple events like $P(X_n/n>x)$ can be computed more or less explicitly. It is a worthwhile idea to try to show the convergence from the definition, that is, by showing the convergence of the distribution function. This is what we do below:
\begin{align}
F_{\frac{X_n}{n}}(t)&=1-P(X_n>tn)\\
&= 1-\left(1-\frac{5}{n}\right)^{tn}\\
&\to 1-e^{-5t}.
\end{align}
This shows that the $\frac{X_n}{n}$ converges to $Exp(5)$ in distribution. (Note that the second equality above is not true, one should be honestly using $\lceil tn \rceil$ but it doesn't matter in the limit, so I cheated a little bit.)
As you mentioned CLT, the usual way of proving CLT is via analyzing the characteristic functions. It is not hard to write the characteristic function of the geometric distribution and showing that the limit indeed is the characteristic function of the exponential distribution. This would give an alternate proof that would be very much in the spirit of (the proof of) CLT.
A: Using MGF you get
$$MGF\Big(\frac{x_n}{n}\Big)=\frac{\frac{5}{n}}{1-\Big(1-\frac{5}{n}\Big)e^{\frac{t}{n}}}\xrightarrow{n\to \infty}=\frac{0}{0}\xrightarrow{H}\frac{5}{5e^{\frac{t}{n}}-\Big(1-\frac{t}{n}\Big)e^{\frac{t}{n}}t}=\frac{5}{5-t}$$
Thus
$$\frac{x_n}{n}\xrightarrow{d}Exp(5)$$
Because we recognize the MGF of an exponential distribution with parameter $\lambda=5$
