Let $\mu$ be a probability measure on $\mathbb R$ and define $$ \alpha_n = \int_{\mathbb R} x^n\,\text d\mu(x) $$ to be the $n$th moment. Suppose that $$ \sum_{n=0}^\infty \frac{\alpha_nt^n}{n!} < \infty $$ for $|t| < \delta$ with $\delta > 0$ being small.
Let $$ \beta_n = \int|x|^n\,\text d\mu(x) $$ be the $n$th absolute moment. Can I conclude that there is some $\varepsilon > 0$ such that $$ |t| < \varepsilon \stackrel ?\implies \sum_{n=0}^\infty \frac{\beta_n t^n}{n!} < \infty $$ ?
I can show that for $|t|$ sufficiently small $\frac{\beta_n t^n}{n!}\to 0$, but I'm not sure how to show convergence or how to come up with a counterexample.
It is well known that the radius of convergence of a Power series $\sum_n a_n z^n$ is $R = \sup \{ r > 0 \colon ( r^n a_n)_n \text{ bounded } \}$. The proof of this uses that if $|z| < r$ and $|a_n r^n| \leq C$, then $\sum_n |a_n z^n|\leq C \sum |(z/r)|^n <\infty$, by comparing with the geometric series.
Since you showed that $\beta_n t^n /n! \to 0$ (you might want to post the proof of this), and since convergent sequences are bounded, this shows that the power series $\sum_n \beta_n t^n/n!$ has positive radius of convergence.