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Theorem: Let $\sum_{n=0}^{\infty}a_nz^n$ be a power series with radius of convergence $R$ and $f:U\to \mathbb{C}$, \begin{equation}f(z)=\sum_{n=0}^{\infty}a_nz^n\end{equation} where $U=B(0,r)$, $0<r<R$. Then $f$ is holomorphic at $0$ and \begin{equation}f^{\prime}(z)=\sum_{n=1}^{\infty}na_nz^{n-1} \end{equation}

Question: I have been reading the proof of this from here pg 80 and I don't understand how the last two lines in that page are derived.


After we show that both series have the same radius of convergence $R$ let $g:U\to \mathbb{C}$, \begin{equation}g(z)=\sum_{n=1}^{\infty}na_nz^{n-1} \end{equation} Choose $h$ with $0<\left|h\right|<r-\left|z\right|$. Then $z+h\in U$. If $u_n(h)=a_n\frac{(z+h)^n-z^n}{h}-na_nz^{n-1}$ then, \begin{equation}u_n(h)=a_n\frac{(z+h)^n-z^n}{h}-na_nz^{n-1}=a_n\sum_{k=0}^{n-1}(z+h)^{n-k-1}z^k-nz^{n-1} \end{equation} Since $\left|z\right|<r$ and $\left|z+h\right|<r$, \begin{equation}\left|u_n(h)\right|\le\left|a_n\right|\sum_{k=0}^{n-1}r^{n-1}+nr^{n-1}=2n\left|a_n\right|r^{n-1} \end{equation} Since the radius of convergence of series $\sum_{n=1}^{\infty}na_nz^{n-1}$ is $R>r$ we have that $\exists N\in \mathbb{N}$ so that \begin{equation}\left|\sum_{n=N}^{\infty}u_n(h)\right|\le \sum_{n=N}^{\infty}\left|u_n(h)\right| \le 2\sum_{n=N}^{\infty}n\left|a_n\right|r^{n-1}<\epsilon \end{equation}

I don't understand how the bold face sentence implies the last inequality. I know that $\sum_{n=N}^{\infty}n\left|a_n\right|r^{n-1}$ converges but how is it less than an arbitary $\epsilon$?.

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up vote 1 down vote accepted

The series $\sum_{n\geqslant 1}n|a_n|r^{n-1}$ is convergent, as $r<\frac{r+R}2$ and $\left\{n|a_n|\left(\frac{r+R}2\right)^{n-1}\right\}$ is bounded. If $\sum_{n\geqslant 1}a_n$ is convergent (to $s$), then the sequence $\left\{s_n:=\sum_{j=1}^na_j\right\}$ is convergent, so $\sum_{j=n}^{+\infty}a_j=s-s_n\to 0$. Now we use the definition of convergence.

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Or we can use the Cauchy's convergence test and then take the limit $n\to \infty$ right? – Nameless Nov 6 '12 at 20:00
Yes (that quite the same approach). – Davide Giraudo Nov 6 '12 at 20:03
One other small question. Is that called Abel's Theorem? – Nameless Nov 6 '12 at 20:04
I think yes, even if Abel has done a lot of other things in this area. – Davide Giraudo Nov 6 '12 at 22:02

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