Let $X$ be a bounded real random variable. Show that its characteristic function is a power series whose radius of convergence is equal to infinity.
My attempt: let $\varphi $ be its characteristic function.
We have $$\varphi _X(t) = \mathbb{E}[e^{itX}] = \int_{\mathbb{R}}e^{itX}\mathbb{P}_X(dx) =\int_{\mathbb{R}}\sum_{n \in \mathbb{N}}\frac{(itx)^n}{n!}\mathbb{P}_X(dx) $$ As $X$ is bounded, by the Beppo-Levi theorem, we can interchange the integral and the sum: $$ \int_{\mathbb{R}}\sum_{n \in \mathbb{N}}\frac{(itx)^n}{n!}\mathbb{P}_X(dx) = \sum_{n \in \mathbb{N}} \frac{i^nt^n}{n!}\int _{\mathbb{R}}x^n \mathbb{P}_X(dx) = x \sum_{n \in \mathbb{N}} \frac{(xit)^n }{n! (n+1)}$$ And I am unsure if what I did is sufficient and even correct. Any ideas?
The justification of the exchange of the integral and the series follows from the fact that $$\int_{\mathbb{R}}\sum_{n \in \mathbb{N}} \left\lvert\frac{(itx)^n}{n!} \right \rvert\mathbb{P}_X(dx)=\int_{\mathbb{R}}\sum_{n \in \mathbb{N}} \frac{\left\lvert t\right \rvert^n}{n!}\left\lvert x\right\rvert^n \mathbb{P}_X(dx)$$ and if $X$ is bounded by $M$, then $$\int_{\mathbb{R}}\sum_{n \in \mathbb{N}} \frac{\left\lvert t\right \rvert^n}{n!}\left\lvert x\right\rvert^n \mathbb{P}_X(dx)=\int_{ [-M,M ]}\sum_{n \in \mathbb{N}} \frac{\left\lvert t\right \rvert^n}{n!}\left\lvert x\right\rvert^n \mathbb{P}_X(dx) \leqslant \int_{ [-M,M ]}\sum_{n \in \mathbb{N}} \frac{\left\lvert t\right \rvert^n}{n!}M^n \mathbb{P}_X(dx)\leqslant \sum_{n \in \mathbb{N}} \frac{\left\lvert t\right \rvert^n}{n!}M^n,$$ which is finite. This leads to $$ \varphi_X(t)=\sum_{n=0}^{+\infty}a_nt^n,\mbox{ where }a_n :=\frac{i^n}{n!}\mathbb E\left[X^n\right] . $$ Since $\left \lvert a_n \right\rvert\leqslant M^n/n!$, the power series has an infinite radius of convergence.