Using results
we have
\begin{equation}
\begin{split}
\mathbb{E} e^{i t G}
& = \sum_{n \in 1, 2, 3, \ldots} e^{i t n} \mathbb{P}(G = n) \\
& = \sum_{n \in 1, 2, 3, \ldots} e^{i t n} \theta (1 - \theta)^{n - 1} \\
& = \theta e^{i t} \sum_{n \in 1, 2, 3, \ldots} e^{i t (n - 1)} (1 - \theta)^{n - 1} \\
& = \theta e^{i t} \sum_{n \in \mathbb{N}} e^{i t n} (1 - \theta)^n \\
& = \theta e^{i t} \sum_{n \in \mathbb{N}} (e^{i t} (1 - \theta))^n \\
& = \theta e^{i t} \frac{1}{1 - e^{i t} (1 - \theta)} \\
& = \theta e^{i t} \frac{1}{1 - e^{i t} (1 - \theta)} \frac{e^{- i t}}{e^{- i t}} \\
& = \frac{\theta}{e^{- i t} - (1 - \theta)} \\
\end{split}
\end{equation}
$\square$