# The coefficient of a power series

Assume there has the power series $\sum_{n=1}^{\infty}a_nx^n$,the convergent radius of it $r>0$,prove:

If $a_1\neq 0$,and in a neighborhood of zero point,$|\sum_{n=1}^{\infty}a_nx^n|\geq|a_1||x|-2x^2$,then

$|a_2|\leq2$

It is a problem in my exercise book, I don't know how to start.

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To begin with, can you show that $g(x)=\sum\limits_{n\geqslant2}a_nx^{n-2}$ is such that $|g(x)|\leqslant2$ for $|x|$ small enough? –  Did Jul 25 '12 at 6:59
@did Since $\sum_{n=1}^{\infty}a_nx^n\to0$ when $x\to0$,I think it is right. –  D-Hunter Jul 25 '12 at 7:02
This is unrelated, please read again the result I propose to prove (in particular, $g(x)$ does not depend on $a_1$). And note that this result is wrong in general hence you must use a hypothesis you did not use so far. –  Did Jul 25 '12 at 7:04
Try figuring it out with $a_n = 0$ for $n>2$ first. Remember you are working in $\mathbb{R}$, and you can choose the sign of $x$ to suit your needs. –  copper.hat Jul 25 '12 at 7:35
89085731: Any progress? –  Did Jul 26 '12 at 12:01

If $a_2=0$, there is nothing to prove, hence from now on we assume that $a_2\ne0$.

Assume first that $a_n=0$ for every $n\geqslant3$ (as @copper.hat suggested). Then the hypothesis is that $|a_1+a_2x|\geqslant|a_1|-2|x|$ for every $|x|$ small enough, say $|x|\leqslant x_0$. Choose $x$ such that $x\ne0$, $|x|\leqslant x_0$, $a_1a_2x\leqslant0$, and $|a_2|\,|x|\leqslant |a_1|$. Then $|a_1|-|a_2|\,|x|=|a_1+a_2x|\geqslant|a_1|-2|x|$ and $x\ne0$, hence $|a_2|\leqslant2$.

In the general case, consider $A(x)=a_2+\sum\limits_{n=3}^{+\infty}a_nx^{n-2}$. Since the series $\sum\limits_na_nx^n$ has a positive radius of convergence, the function $A$ is continuous at $0$ and $A(0)=a_2\ne0$, hence $A(x)\ne0$, $|A(x)|\leqslant2|a_2|$, and $A(x)$ has the sign of $a_2$, for every $|x|$ small enough, say $|x|\leqslant x_0$. One can, and we will, assume without loss of generality that, furthermore, $|a_1+A(x)x|\geqslant|a_1|-2|x|$ for every $|x|\leqslant x_0$. Choose $x$ such that $x\ne0$, $|x|\leqslant x_0$, $a_1A(x)x\leqslant0$ and $2|a_2|\,|x|\leqslant |a_1|$. Then $|a_1|-|A(x)|\,|x|=|a_1+A(x)x|\geqslant|a_1|-2|x|$ and $x\ne0$, hence $|A(x)|\leqslant2$. One can choose $x$ with all these properties as close to $0$ as desired, hence, by the continuity of $A$ at $0$, $|a_2|\leqslant2$.

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Starting from the fact that $|\sum_{n=1}^{\infty} a_{n}x^n | \leq \sum_{n=1}^{\infty}|a_{n}||x^n|$, we have

$$|\sum_{n=1}^{\infty} a_{n}x^n| \geq |a_{1}||x|-2 |x^2| \Rightarrow \sum_{n=2}^{\infty} |a_{n}| |x^n| \geq -2 x^2 \Rightarrow (|a_2|+2)x^2 + \sum_{n=3}^{\infty} |a_{n}||x^n| \geq 0$$

$$\Rightarrow (|a_2|+2+|a_3| |x| + |a_4| x^2+\dots)x^2 \geq 0 \Rightarrow (|a_2|+2+|a_3||x|+a_4 x^2 +\dots) \geq 0 \Rightarrow$$ Taking the limit of the last inequality as $x$ goes to 0 and solving resulting inequality gives the desired result $$|a_2| \leq 2$$

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-1. To reach the conclusion that $|a_n|+2\geqslant0$ is not what I would call a tour de force. –  Did Jul 25 '12 at 11:14
Sorry but the new version is still reaching the earth-shattering conclusion that $|a_2|+2\geqslant0$. (Do you really read the comments made on your answers?) –  Did Jul 25 '12 at 16:21
@did:You were not able to give a single hint to the problem and now you are commenting on solution!!. –  Mhenni Benghorbal Jul 26 '12 at 3:00
The first part of your comment is factually false. The second part is factually true (with or without the exclamation marks). Both parts have nothing to do with the fact that your solution is wrong and that you do not see fit to signal this, even when it is mentioned to you. –  Did Jul 26 '12 at 7:16
@Downvoters: it is an old game. Just leave my answers alone. Other people can read and understand what you could not. –  Mhenni Benghorbal Dec 9 at 2:00