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Let $f(z) = \sum a_n z^n$ be a power series with radius of convergence $R$. How do we show that $f$ is analytic in the circular region of radius $R$?

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You know that the power series itself converges inside the radius of convergence. What can you say about the formal derivative of that power series? If it converges, the term by term derivative is a valid differentiation of the function given by the power series. –  hardmath Aug 26 '12 at 0:29
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isn't there any theorem something like this. –  Monkey D. Luffy Aug 26 '12 at 0:33
    
Sure, are you studying a complex variables text? The proof is not hard --- you just need to control for the growth of the exponents $n$ that you bring down in term by term derivatives. –  hardmath Aug 26 '12 at 0:35
    
yeah I'm studying it ... I have proof in my book. I don't understand how it relates R so that |(f(z+h) - f(z))/h - f'(z))| = 0. Besides book is old ... I was looking for it somewhere online or ... on any ebook. –  Monkey D. Luffy Aug 26 '12 at 0:37
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3 Answers

up vote 8 down vote accepted

Lemma Let $R$ be the radius of convergence of $f(z) = \sum_{n=0}^{\infty} a_nz^n$. Let $g(z) = \sum_{n=1}^{\infty} na_nz^{n-1}$.

Then the radius of convergence of $g(z)$ is $R$.

Proof: Let $R'$ be radius of convergence of $g(z)$. Since $|a_nz^n| \le |na_nz^n|$, $R' \le R$. So it suffices to prove $R \le R'$.

Let $z$ be such that $|z| < R$. Choose $r$ such that $|z| < r < R$. Let $\rho = \frac{|z|}{r}$.

Since $\sum_{n=1}^{\infty} a_nr^{n-1}$ converges, there exists $M > 0$ such that $|a_nr^{n-1}| \le M$ for all $n \ge 1$.

Then

$|na_nz^{n-1}| = n|a_n|r^{n-1}\rho^{n-1} \le nM\rho^{n-1}$

Since $0 \le \rho < 1$, $\sum_{n=1}^{\infty} nM \rho^{n-1}$ converges. Hence $\sum_{n=1}^{\infty} na_nz^{n-1}$ converges. Hence $R \le R'$. QED

Proposition Let $R$ be the radius of convergence of $f(z) = \sum_{n=0}^{\infty} a_nz^n$. Let $g(z) = \sum_{n=1}^{\infty} na_nz^{n-1}$.

Then $f'(z) = g(z)$ for every $|z| < R$.

Proof: Let $z$ be such that $|z| < R$. Choose $r$ such that $|z| < r < R$. Let $h$ be such that $0 < |h| \le r - |z|$

Consider $\frac{f(z+h) - f(z)}{h} = \sum_{n=0}^{\infty} a_n\frac{(z + h)^n - z^n}{h}$

By the formula $x^n - y^n = (x - y)(x^{n-1} + yx^{n-2} +\cdots+ y^{n-2}x + y^{n-1})$,

$|a_n\frac{(z + h)^n - z^n}{h}| = |a_n||(z + h)^{n-1} + z(z+h)^{n-2}+\cdots+ z^{n-2}(z+h) + z^{n-1}) \le n|a_n|r^{n-1}$

Define $\psi_n(h) = a_n\frac{(z + h)^n - z^n}{h}$ for $0 < |h| \le r - |z|$

Define $\psi_n(0) = n a_n z^{n-1}$.

Then $\psi_n(h)$ is continuous in $|h| \le r - |z|$. Since $\sum_{n=1}^{\infty} n|a_n|r^{n-1}$ converges by the lemma, $\Psi(h) = \sum_{n=1}^{\infty} \psi_n(h)$ converges uniformly in $|h| \le r - |z|$.

Hence $\Psi(h)$ is continuous in $|h| \le r - |z|$. In particular $\lim_{h \to 0} \Psi(h) = \Psi(0)$.

This implies $f'(z) = g(z)$.

QED

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I think all you need is

Theorem: A power series converges uniformly inside its convergence interval

Proof: Follows at once from Weierstrass M-test: let $\,r>0\,$ be the convergence radius of the power series $$\sum_{n=1}^\infty a_n(z-a)^n$$ and let $\,0<\rho<r\,$, then: $$|z-a|<\rho\Longrightarrow |a_n(z-a)^n|\leq |a_n|\rho^n=:M_n$$ and since the series $\,\displaystyle{\sum_{m=1}^\infty M_n}\,$ converges (absolutely, of course) , we're done.

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I don't see why the series $\sum_{m=1}^\infty M_n$ should converge. And a power series converges uniformly on closed balls contained in the open ball of radius $r$, but I don't think it necessarily converges uniformly on the entire open ball. –  Keenan Kidwell Aug 26 '12 at 3:54
    
1) A power series converges always for values within its interval of convergence. 2) The series converges on the compact set $\,[\rho-\epsilon,\rho+\epsilon]\,$ , with $\,\epsilon\,$, small enough as to have $\,\rho+\epsilon<r\,$ –  DonAntonio Aug 26 '12 at 4:04
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I will show another proof.

Proposition Let $R$ be the radius of convergence of $f(z) = \sum_{n=0}^{\infty} a_nz^n$. Let $g(z) = \sum_{n=1}^{\infty} na_nz^{n-1}$.

Then $f'(z) = g(z)$ for every $|z| < R$.

Proof: By the lemma of my previous answer, the radius of convergence of $g(z)$ is $R$. Let $z$ be such that $|z| < R$. Since $\sum_{n=1}^{\infty} na_nz^{n-1}$ converges uniformly in every compact subset of $\{z; |z| < R\}$,

$\int_{0}^{z} g(\zeta) d\zeta = \sum_{n=1}^{\infty} \int_{0}^{z} na_n\zeta^{n-1} d\zeta = \sum_{n=1}^{\infty} a_nz^n = f(z) - a_0$,

where the integral path is any smooth curve inside the domain $\{z; |z| < R\}$ starting from $0$ to $z$.

Hence $f'(z) = g(z)$.

QED

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It's neither polite nor used to write more than 1 answer to the same question, so I'll tell you what they told me when I did this: add this stuff (which, btw, is very nice) to your original answer. –  DonAntonio Aug 27 '12 at 7:55
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@DonAntonio "It's neither polite nor used to write more than 1 answer to the same question" Why it's not polite to write more than one answer to the same question? –  Makoto Kato Aug 27 '12 at 11:28
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