Sorry for the missing constraints. As you guessed, $X_1, \dots, X_n$ are i.i.d. copies of $X$, and $X$ is a nonnegative random variable. Thank you very much.

because $\psi_n\to\psi$ locally uniformly on $|z|<1$, so the derivatives at $z=0$ converge. -> why this is true? Would be appreciated for references.

Thank you very much. I think I've seen the convergence theorem related to uniformly integrable random variables. I'm asking this just to make sure, the uniformly integrability of $Y_n$ does not effected by the uniformly integrability of $X_n$? In fact, $X_n$ is not uniformly integrable since $EX = \infty$ in my case.

You're right, they take values in 0.1,2,... Thank you very much!

Thank you for a great reference. I've seen a hint attached to Exercise 24, and it said to prove that $\bar{X}_n$ is unbounded almost surely. But what if $\bar{X}_n$ is always bounded? In my case, $X_n$ is a bounded random variable such that $0 \leq X_n \leq A_n$, with $A_n\to\infty$ as $n\to\infty$. Still not true?