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. $\square$
