I've seen that the moment generating function of the Beta Distribution is the following:
$${\displaystyle 1+\sum _{k=1}^{\infty }\left(\prod _{r=0}^{k-1}{\frac {\alpha +r}{\alpha +\beta +r}}\right){\frac {t^{k}}{k!}}}$$
However, nowhere have I seen it specified for what values of $t$ this moment generating function is valid. Is it only valid for positive $t$? All real numbers $t$? I'm not sure, nor do I know how to show what values it would be valid for.
Using the Cauchy Root Test we can establish that the radius of convergence of this moment generating function is infinite, and so therefore it is defined for all $t \in \mathbb C$.
Recall, that the Cauchy Root Test implies that the radius of convergence, $R$, of a power series $A(z) = \sum_{n=0}^\infty a_n z^n$ satisfies $$ \frac{1}{R} = \limsup_{n \rightarrow \infty} |a_n|^{\frac1n}.$$
So applying this to the Moment generating function of a Beta distribution with $\alpha, \beta > 0$ \begin{align*} C & = \limsup_{n \rightarrow \infty}|a_n|^{\frac1n} \\ & = \limsup_{n \rightarrow \infty} \left| \frac{1}{n!} \prod_{r=0}^{n-1} \frac{\alpha + r}{\alpha + \beta + r} \right|^{\frac1n} \\ & \leq \limsup_{n \rightarrow \infty} \left| \frac{1}{n!} \prod_{r=0}^{n-1} 1\right|^{\frac1n} \\ & = \limsup_{n \rightarrow \infty} \left| \frac{1}{n!}\right|^{\frac1n} \\ & \leq \limsup_{n\rightarrow \infty} \left( \frac{1}{n^n} \right)^{\frac1n} \\ & = \limsup_{n \rightarrow \infty} (n^{-n})^{\frac1n}\\ & = \limsup_{n\rightarrow \infty} n^{-1}\\ & = 0. \end{align*} It follows (from the Cauchy Root Test) that the radius of convergence is $R = \infty$.