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.