# The proof of Abel's limit theorem

I came up with a question for a proof of Abel's limit theorem (P.41 Complex Analysis by Ahalfors). Help from those who have this book is appreciated.

The theorem states that if a power series $a_0+a_1 z+ a_2 z^2+...$ has the convergence radius $R=1$ and converges at $z=1$, then a function $f(z)=a_0+a_1 z+ a_2 z^2+...$approaches to $f(1)$ as $z\to 1$ in a such way that $|1-z|/(1-|z|)$ is bounded.

In the proof the author defines a partial sum $s_n=a_0+a_1+a_2+...$ and say "$s_n z^n \to 0$". My question is that: 1) "$s_n z^n \to 0$" as $n \to 0$ or $z \to 1$? 2) Why does it tend to $0$?

I don't have the book, but let me guess: It is common to simplify the proof by assuming $f(1)=0$ (otherwise just subtract a constant from $f$). When that is done, $s_n\to f(1)=0$ as $n\to\infty$. –  Harald Hanche-Olsen Oct 31 '12 at 14:29
@HaraldHanche-Olsen Thanks for comment. Yes. $s_n \to 0$ but it says "$s_n z^n \to 0$. –  Tengu Oct 31 '12 at 14:39
But $\lvert z\rvert<1$, so $s_n\to0$ implies $s_nz^n\to0$. –  Harald Hanche-Olsen Oct 31 '12 at 15:21
I have the book, but not right here, and I agree it should be $n\to\infty$. You don't even have to subtract a constant, since $(s_n)$ is bounded and $|z|<1$, you get that $s_n z^n \to 0$. –  Lukas Geyer Oct 31 '12 at 15:23
@HaraldHanche-Olsen, Lukas I didn't take into consideration that $|z|<1$. Thank you very much! –  Tengu Nov 1 '12 at 1:55