# Let $f($z$)= \sum _{n=0}^{\infty } a_nz^n$ be the power series expansion of f about $0$. Prove that $|a_n| \le (n+1)(1+1/n)^{n} < e(n+1)$

Suppose that $f$ is analytic in the unit disc D = {$z \in \mathbb{C}$ : |$z$| < 1} and $|$f($z$)$| \le 1/(1-|$z$|)$ for all $z\in D$.

Let $f($z$)= \sum _{n=0}^{\infty } a_nz^n$ be the power series expansion of f about $0$.

Prove that $$|a_n| \le (n+1)(1+1/n)^{n} < e(n+1)$$

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I've been working on this problem for a while now, but can't seem to get anywhere. I've tried expressing 1/(1-$|$z$|$ as the geometric series $\sum _{n=0}^{\infty} |z|^n$ and writing $|f($z$)|= \sum _{n=0}^{\infty } |a_n||z|^n$ $\le$ $\sum _{n=0}^{\infty} |z|^n$ – matt Oct 14 '12 at 0:32
Also, I know that $$e=\lim_{n \to \infty} (1 + 1/n)^n.$$ My approach was to prove the $\le$ inequality first and then use this fact to show the last inequality – matt Oct 14 '12 at 0:33
Please avoid using $$ in titles. – Pedro Tamaroff Oct 14 '12 at 0:43 k, thanks. sorry! – matt Oct 14 '12 at 0:45 The second inequality is really easy, use (1+\frac{1}{n})^n <e<(1+\frac{1}{n})^{n+1} – Alex Oct 14 '12 at 0:50 ## 2 Answers Recall that Cauchy integral formula gives us$$ a_n= \frac{1}{2\pi i}\int\limits_{\partial B(0,r)}\frac{f(z)}{z^{n+1}}dz \frac{1}{2\pi i}\int\limits_{0}^{2\pi}\frac{f(r e^{it})}{(re^{it})^{n+1}}d(r e^{it})= \frac{1}{2\pi}\int\limits_{0}^{2\pi}\frac{f(r e^{it})}{r^n e^{int}}dt $$hence$$ |a_n|\leq \frac{1}{2\pi}\int\limits_{0}^{2\pi}\frac{|f(r e^{it})|}{r^n |e^{int}|}dt\leq \frac{1}{2\pi}\int\limits_{0}^{2\pi}\frac{dt}{r^n (1-|r e^{it}|)}= \frac{1}{r^n (1-r)} $$As the consequence$$ a_n\leq\min\limits_{r\in(0,1)}\frac{1}{r^n (1-r)}=(n+1)\left(1+\frac{1}{n}\right)^n $$I suggest you to fill the gaps. - thanks for the help! – matt Oct 14 '12 at 1:32 sorry, i'm still not sure how you get the equality on the right hand side of$$a_n\leq\min\limits_{r\in(0,1)}\frac{1}{r^n (1-r)}=(n+1)\left(1+\frac{1}{n}\right)^n – matt Oct 14 '12 at 1:50
use calculus to find point of minimum. Find derivative, find stationary points, substitute into the function – Norbert Oct 14 '12 at 1:59
thanks again :D – matt Oct 14 '12 at 2:01

Here are some hints.

1. Cauchy's theorem can help you recover $a_n$ from $f$.
2. Use the inequality $|\int f| \leq \int |f|$
3. What is the minimum of $\frac{1}{z^{n+1}(1-z)}$ when $z\in (0,1)$?
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thanks for the help! – matt Oct 14 '12 at 1:32