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I am busy reading John Conway's Functions of One Complex Variable I (2nd ed.). On page 36 I got stuck on the proof part a) of proposition 2.5.

2.5 Proposition. Let $f(z)=\displaystyle\sum_{n=0}^{\infty}a_n(z-a)^n$ have radius of convergence $R>0$. Then:

(a) For each $k\geq 1$ the series $$\sum_{n=k}^{\infty}n(n-1)\dots (n-k+1)a_n(z-a)^{n-k} \tag{2.6}$$ has radius of convergence R.

Proof. Assume that $a=0$.

(a) We first remark that if (a) is proved for $k=1$ then the cases $k=2, \dots$ will follow. In fact, the case $k=2$ can be obtained by applying part (a) for $k=1$ to the series $\sum na_n(z-a)^{n-1}$. We have that $R^{-1}=\lim\sup |a_n|^{1/n}$; we wish to show that $R^{-1}=\lim\sup|na_n|^{1/(n-1)}$. Now it follows from l'Hôpital's rule that $\displaystyle\lim_{n\to\infty}\frac{\log n}{n-1}=0$, so that $\displaystyle\lim_{n\to\infty}n^{1/(n-1)}=1$. The result will follow from Exercise 2 if it can be shown that $\lim\sup |a_n|^{1/(n-1)}=R^{-1}$.

Let $(R')^{-1}=\lim\sup|a_n|^{1/(n-1)}$; then $R'$ is the radius of convergence of $\displaystyle\sum_1^{\infty}a_nz^{n-1}=\sum_0^{\infty}a_{n+1}z^n$. Notice that $z\sum a_{n+1}z^n+a_0=\sum a_nz^n$; hence if $|z|<R'$ then $\sum|a_nz^n|=|a_0|+|z|\sum|a_{n+1}z^n|<\infty$. This gives $R'\leq R$.

I understand everything up to here; the last part is because when $|z|<R'$, $\sum |a_{n+1}z^n|$ converges.

If $|z|<R$ and $z≠0$ then $\sum|a_nz^n|<\infty$ and $\displaystyle\sum|a_{n+1}z^n|=\frac 1 {|z|}$. $\displaystyle\sum|a_nz^n|+\frac 1 {|z|}|a_0|<\infty$, so that $R\leq R'$. This gives that $R=R'$ and completes the proof of part (a).

Please can you explain why $\displaystyle\sum|a_{n+1}z^n|=\frac 1 {|z|}$? I also don't see why the implication in the following sentence is true.

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2 Answers 2

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I think you're confused by the simple fact that the point after $\frac 1 {|z|}$ means 'product' and not 'end of sentence'.

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  • $\begingroup$ Why is there not a minus sign in front of $\frac 1 {|z|}|a_0|$? $\endgroup$
    – ahorn
    Sep 3, 2016 at 14:20
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Observe that

$$\sum_{n=0}^\infty \left|a_{n+1}z^n\right|=\frac1{|z|}\sum_{n=0}^\infty|a_{n+1}z^{n+1}|=\frac1{|z|}\sum_{n=0}^\infty|a_nz^n|-\frac{|a_0|}{|z|}$$

and now just use

$$|a\pm b|\le|a|+|b|\;,\;\;\;\forall\,a,b\in\Bbb C$$

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