Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

up vote 3 down vote accepted

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).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.