Generating functions and Weak Convergence

Question

I'm trying to prove an inequality involving a sequence of probability distributions $P_n(k)$ and their generating functions $F_n(z)$, where $n = 1,2,3...$ and $F(z) = \sum_{k=0}^\infty P(k)z^k$ and $F_n(z) = \sum_{k=0}^\infty P_n(k)z^k$ .
Given that $\{P_n(k)\}$ converges weakly to $P(k)$, the inequality I'm trying to prove is $$|F_n(z) - F(z)| \leq \sum_{k=0}^K|P_n(k) - P(k)| + \sum_{k=K+1}^\infty|z|^k$$ for any positive integer $K$. Using triangle inequality I have $$|F_n(z) - F(z)| = |\sum_{k=0}^\infty(P_n(k)z^k - P(k)z^k)| $$ $$\leq \sum_{k=0}^\infty |z^k(P_n(k) - P(k))| $$ $$ = \sum_{k=0}^\infty |z|^k |P_n(k) - P(k)|$$ This is equal to $$\sum_{k=0}^K |P_n(k) - P(k)| |z|^k + \sum_{k=K+1}^\infty |P_n(k)-P(k)||z|^k$$ But I don't know what to do from here or how to use the fact that $\lim_{n\rightarrow \infty}P_n(k) = P(k)$.

Answer 1

The inequality you are trying to prove does not require weak convergence. In the last step just note that $|P_n (k)-P(k)|\leq 1$ and you are done.

Answer 2

The main idea in the proof needs to be that weak convergence of the sequence $P_n$ implies that there exists some $L > 0$ with

$$\sum_{l \ge L} P_n(l) < 1/2$$

uniformly in $n$. This is formally called "tightness" of the sequence of probability measures. If you have this result available, the proof is more or less automatic.

Edit: Something is confusing me here. As it stands, the result is trivial. Both $F_n(z)$ and $F(z)$ are between $0$ and $1$, so their difference is as well. Therefore the result needs to be shown only for $|z| < 1$, as the right hand side of the inequality is infinite otherwise. But then the triangle inequality directly gives the result, because $|P_n(k) - P(k)| \le 1$ always as well. So no weak convergence is needed.