# Convergence of probability generating function implies convergence in distribution

Let $\{X_n\}_{n\in \mathbb{N}}$ be a sequence of nonnegative discrete random variables, and $X$ be a nonnegative discrete random variable. The probability generating function $\psi_X(z)=\sum_{k=0}^\infty z^k \mathbb{P}(X=k)$ is convergent when $z \leq 1$.

Now assume that $\psi_{X_n}(z) \to \psi_X(z)$ for all $z\leq 1$. Does this implies that $X_n\to X$ in distribution?

Here, theorem 2.10 says that if we can find a neighborhood of 1 such that $\psi_{X_n}(z)\to \psi_{X}(z)$ for all $z$ in that neighborhood, $X_n\to X$ in distribution. However, in my case, $\psi_X(z)$ is convergent when $z\leq 1$, so we cannot find a neighborhood of 1 satisfy the condition of theorem 2.10. Does this mean that $X_n$ may not converge to $X$ even if the PGF converges?
• Yes, this follows, because $\psi_n\to\psi$ locally uniformly on $|z|<1$, so the derivatives at $z=0$ converge. (You don't say this very explicitly, but I'm assuming your rv's take values in $0,1,2, \ldots$.) – Christian Remling Jan 28 '17 at 18:08
• 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. – user3141978 Feb 15 '17 at 17:25