# Evaluating the convergence of a complex series

This is a part of the proof on a textbook of the fact that a power series that converges on an open disc defines an analytic function.

First note this inequality about a real series: for any $N,N^\prime\in\mathbb{N}$, $$\sum_{q=0}^{N^\prime}\sum_{p=0}^{N}\binom{p+q}{p}|a_{p+q}||b|^p|\zeta|^q\le\sum_{n=0}^\infty\sum_{p+q=n}\binom{p+q}{p}|a_{p+q}||b|^p|\zeta|^q=\sum_{n=0}^\infty|a_n|(|b|+|\zeta|)^n < \infty,$$ where $|b|$ is less than the radius $r$ of the convergence of the series $\sum_{n=0}^\infty a_nz^n$, and $0 < |\zeta| < r - |b|$. Then, consider the complex series $\sum_{q=0}^{N}\sum_{p=0}^{N^\prime}\binom{p+q}{p}a_{p+q}b^p\zeta^q$, and decompose it into two like this: $$\left(\sum_{p+q\le N}+\sum_\text{other pairs}\right)\binom{p+q}{p}a_{p+q}b^p\zeta^q.$$ Let $R$ be the second term. Here the textbook goes on to say that from the first inequality, for any given $\epsilon>0$, $|R|<\epsilon$ holds for sufficiently large $N$.

Why the last statement holds?

-

The statement follows from

$$\sum_{n=0}^\infty\sum_{p+q=n}\binom{p+q}{p}|a_{p+q}||b|^p|\zeta|^q < \infty.\tag{1}$$

In particular, the series converges.

We have

$$\sum_{p+q\le N}\binom{p+q}{p}a_{p+q}b^p\zeta^q = \sum_{n=0}^N\sum_{p+q=n}\binom{p+q}{p}|a_{p+q}||b|^p|\zeta|^q,$$

and the "other terms" part, $R$, satisfies

$$|R| = \left|\sum_\text{other pairs}\binom{p+q}{p}a_{p+q}b^p\zeta^q\right| \leq \sum_{n=N+1}^\infty\sum_{p+q=n}\binom{p+q}{p}|a_{p+q}||b|^p|\zeta|^q$$

by just filling in the extra terms up to $\infty$. The statement $|R| \to 0$ (the epsilon thing) is just a restatement of the fact that $(1)$ converges (that the tail of the series gets arbitrarily small).

-