# Question about a statement in a proof about complex power series

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 $$|z| then $$\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 $$|z|, $$\sum |a_{n+1}z^n|$$ converges.

If $$|z| and $$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.

I think you're confused by the simple fact that the point after $\frac 1 {|z|}$ means 'product' and not 'end of sentence'.
• Why is there not a minus sign in front of $\frac 1 {|z|}|a_0|$? Sep 3, 2016 at 14:20
$$\sum_{n=0}^\infty \left|a_{n+1}z^n\right|=\frac1{|z|}\sum_{n=0}^\infty|a_{n+1}z^{n+1}|=\frac1{|z|}\sum_{n=0}^\infty|a_nz^n|-\frac{|a_0|}{|z|}$$
$$|a\pm b|\le|a|+|b|\;,\;\;\;\forall\,a,b\in\Bbb C$$