Confusion about the expecation of infinite summations

I am confused about the expectation and Taylor series. For example, I assume that the random variable $X$ is bounded, then appling Taylor expansion can give \begin{align} \mathbb{E}\left[e^{X}\right] = \mathbb{E}\left[\sum_{k=0}^{\infty} \frac{X^k}{k!}\right]. \end{align} Now, is the following equality correct? \begin{align} \mathbb{E}\left[\sum_{k=0}^{\infty} \frac{X^k}{k!}\right]=\sum_{k=0}^{\infty}\frac{1}{k!}\mathbb{E}\left[{X^k}\right]. \end{align} However, if I assume $\mathbb{E}\left[{X^k}\right]=k!$, then the above series will diverge, which contradicts with the fact that $\mathbb{E}\left[e^{X}\right]$ exists. Which step causes this contradiction?

Solution 1:

If $|X| \leq M$ then $\sum\limits_{k=1}^{n}|\frac {X^{k}} {k!}| \leq e^{M}$ so DCT can be used to justify interchange of sum and integral. This shows that you cannot have bounded r.v. $X$ with $EX^{k}=k!$. In fact, $E|X|^{k} \leq M^{k}$ and $\frac {M^{k}} {k!} \to 0$ as $k \to \infty$.