I am very puzzled by the following question: $\forall x \in \mathbb{R}$, $e^{ax}$can be expressed as
$$e^{ax}=\sum_{n=0}^{\infty}\frac{a^n}{n!}x^n$$
with an infinite radius of convergence.
Now, let $X_t$ be a Geometric Brownian motion process defined by the SDE:
$$dX_t=\mu X_t dt+\sigma X_t dW_t$$
Then, for a fixed time $t$,by linearity:
$$\mathbb{E}[e^{aX_t}]=\mathbb{E}\left[\sum_{n=0}^{\infty}\frac{a^n}{n!}X_t^n\right]=\sum_{n=0}^{\infty}\frac{a^n}{n!}\mathbb{E}\left[X_t^n\right]=\sum_{n=0}^{\infty}\frac{a^n}{n!}X_0^n e^{\left(n\mu +\frac{n(n-1)}{2}\sigma^2\right) t }$$
But this series is divergent, so what am I missing?