Let $G_n \in \text{Geometric}(\theta_n)$ be a D4001: Geometric random positive integer for each $n \in \mathbb{N}$ such that

(i)	$\lambda \in (0, \infty)$ is a D5407: Positive real number
(ii)	\begin{equation} \lim_{n \to \infty} \theta_n n = \lambda \end{equation}

Let $a \in (0, \infty)$ be a D5407: Positive real number.

For a fixed $n \in \mathbb{N}$, result R4805: Dual probability distribution function for geometric random positive integer shows that one has \begin{equation} \mathbb{P} \left( \frac{G_n}{n} > a \right) = \mathbb{P} ( G_n > n a ) = (1 - \theta_n)^{n a} \end{equation} and therefore \begin{equation} \mathbb{P} \left( \frac{G_n}{n} \leq a \right) = 1 - (1 - \theta_n)^{n a} \end{equation} Since $\lim_{n \to \infty} \theta_n n = \lambda$, then $\lim_{n \to \infty} - \theta_n = 0$ and $\lim_{n \to \infty} - a \theta_n n = - a \lambda$. Therefore, using result R3304: Approximating sequence for the natural exponential function, one has \begin{equation} \lim_{n \to \infty} \mathbb{P} \left( \frac{G_n}{n} \leq a \right) = 1 - \lim_{n \to \infty} (1 - \theta_n)^{n a} = 1 - e^{- \lambda a} \end{equation} This is what was required to be shown. $\square$