Limiting distribution Let $(q_n)_{n>0}$ be a real sequence such that $0<q_n<1$ for all $n>0$ and $\lim_{n\to \infty} q_n = 0$.
For each $n > 0$, let $X_n$ be a random variable, such that
$P[X_n =k]=q_n(1−q_n)^{k−1}, (k=1,2,...)$. 
Prove that the limit distribution of
$\frac{X_n}{\mathbb{E}[X_n]}$
￼is exponential with parameter 1.
I see that $\mathbb{E}[X_n] = \frac{1}{q_n}$ but after that I don't really know where to go from there. Are there any tips please?
 A: Note that for every nonnegative integer $x$, $\mathbb P(X_n\gt x)=(1-q_n)^x$. Furthermore, $Y_n=q_nX_n$ is almost surely nonnegative and, for every nonnegative real number $y$,
$$
\mathbb P(Y_n\gt y)=\mathbb P(X_n\gt x_n)=(1-q_n)^{x_n},
$$
where $x_n$ is the unique nonnegative integer such that $q_nx_n\leqslant y\lt q_n(x_n+1)$. Since $q_n\to0$ and $q_nx_n\to y$, one sees that $(1-q_n)^{x_n}\to\mathrm e^{-y}$. Hence, for every nonnegative real number $y$,
$$
\mathbb P(Y_n\gt y)\to\mathbb P(Y\gt y),
$$
where $Y$ is exponentially distributed with parameter $1$. This proves the claim.
A: First we calculate the characteristic function of $X_n$:
$$\Phi_{X_n}(\xi) = \sum_{k=1}^\infty \underbrace{q_n}_{(q_n-1)+1} \cdot (1-q_n)^{k-1} \cdot e^{\imath \, k \cdot \xi} = - \sum_{k=1}^\infty (1-q_n)^k \cdot (e^{\imath \, \xi})^k+e^{\imath \, \xi} \sum_{k=1}^\infty (1-q_n)^{k-1} \cdot e^{\imath \, (k-1) \cdot \xi} \\ = - \left( \frac{1}{1-(1-q_n) \cdot e^{\imath \, \xi}} - 1 \right) + e^{\imath \, \xi} \cdot \left( \frac{1}{1-(1-q_n) \cdot e^{\imath \, \xi}} \right) \\ = \frac{1}{1-(1-q_n) \cdot e^{\imath \, \xi}} \cdot (-1+(1-(1-q_n) \cdot e^{\imath \, \xi})+e^{\imath \, \xi}) = \frac{q_n \cdot e^{\imath \, \xi}}{1-(1-q_n) \cdot e^{\imath \, \xi}}$$
From this we obtain easily the characteristic function of $Y_n := \frac{X_n}{\mathbb{E}X_n} = q_n \cdot X_n$:
$$\Phi_{Y_n}(\xi) = \Phi_{X_n}(\xi \cdot q_n) = \frac{q_n \cdot e^{\imath \, q_n \cdot \xi}}{1-(1-q_n) \cdot e^{\imath \, q_n \cdot \xi}}$$
Now we let $n \to \infty$ and obtain by applying Bernoulli-Hôpital 
$$ \lim_{n \to \infty} \Phi_{Y_n}(\xi) = \lim_{n \to \infty} \frac{e^{\imath \, q_n \cdot \xi} \cdot (1+q_n \cdot \imath \, \xi)}{-e^{\imath \, q_n \cdot \xi} \cdot (\imath \, \xi \cdot (1-q_n) -1)} = \frac{1}{1-\imath \xi}$$
(since $q_n \to 0$ as $n \to \infty$). Thus we have shown that the characteristic functions converge pointwise to the characteristic function of exponential distribution with parameter 1. By Lévy's continuity theorem we obtain $Y_n \to \text{Exp}(1)$ in distribution.
