The question is from Stein & Shakarchi – Complex Analysis Chapter 2, Exercise 14.
Suppose that $f$ is holomorphic in an open set $\Omega$ containing the closed unit disc, except for a pole at $z_0$ on the unit circle. Show that if $f$ is given by a power series expansion
$$f(z)=\sum^\infty_{n=0}a_n z^n$$
in the unit disc $D_1(0)$, then
$$\lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_0.$$
I solved this problem by using the pole formula
$$f(z)=(z-z_0)^{-m}g_0(z)\implies f^{(n)}(z)=(z-z_0)^{-m-n}g_n(z)$$
where $g_0$ and $g_n$ are holomorphic and not zero at $z_0$, and $m$ is a positive integer. These are defined on $D_{1+\epsilon}(0)\subset\Omega$, which contains $z_0$ and $\epsilon$ is sufficiently small. I've got below :
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}}=\lim_{n\to\infty} \dfrac{\dfrac{f^{(n+1)}(0)}{(n+1)!}}{\dfrac{f^{(n)}(0)}{(n)!}}=\lim_{n\to\infty}\dfrac{1}{n+1}(\dfrac{-m-n}{0-z_0}+H(0))=\dfrac{1}{z_0}.$$
where $H(z)$ is holomorphic in the disc.
Actually, the pole formula I've used appears in Chapter 3 so I should not use this, but I think it is worth to try. Is there something wrong about this solution?
Very thanks.
Best Answer
First note that the proof you gave works perfectly well for $f(z)=\frac{1}{z+1} + \frac{1}{z-1}$. Your functions $g_n$ will not be defined in $D_{1+\epsilon}(0)$, but they are defined on an open set containing $0$, which is enough for your proof to go through. This however would yield a contradiction, since then you would have shown that $-1=\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=1$.
So what is the flaw in your proof? I see two. First, you assumed that none of the $a_n$ are $0$. Indeed if you look at the Taylor series for the function $f$ I defined, you will see that every even term is $0$. Thus the limit doesn't even exist.
Second, I'm not sure how you got that function $H$. Your proof is missing a lot of details, and when I try to reproduce it instead of $H$ I get
$$\frac{g_n'(0)}{g_n(0)}\frac{1}{n+1}$$
If you're looking for solutions, this has been asked before.