Does convergence of all finite moments imply convergence of moment generating function?

Question

Let's say we have a sequence of random variables $X_n$ converges in all finite moments to $X$. In other words, $$ \mathbb{E}(|X_n-X|^k) \to 0, $$ for all $k \in \mathbb N$. Then do we have the convergence of moment generating function, i.e., $$ \mathbb E(e^{t X_n}) \to \mathbb E(E^{t X}) $$ for all $t \in \mathbb R$?

Answer 1

No, this is, in general, not true. Consider for instance random variables $X_n$ such that

$$\mathbb{P}(X_n = n) = e^{-n} \qquad \mathbb{P}(X_n = 0) = 1-e^{-n}$$

and $X:=0$. Then

$$\mathbb{E}(|X_n|^k) = e^{-n} n^k \xrightarrow[]{n \to \infty} 0$$

for all $k \geq 1$, but

$$\mathbb{E}e^{tX_n} = (1-e^{-n}) + e^{-n} e^{tn}$$

does not converge to $1= \mathbb{E}e^{tX}$ for $t \geq 1$.