# An analytic function with a simple pole

Let $f(z)$ be analytic in the disk $|z|<R \ \ \ (R>1)$ except for a simple pole at a point $z_0$, $|z_0|=1$. Consider the expansion $f(z)=a_0+a_1 z+ \cdots$, and show that $$\lim_{n \to \infty} \frac {a_n} {a_{n+1}}=z_0$$

All my attempts failed. I wanted to use the Laurent series at $z_0$ but the problem needs expansion at $0$.

#### Solutions Collecting From Web of "An analytic function with a simple pole"

There is an $a\ne0$ such that the principal part of $f$ at $z_0$ is given by $z\mapsto{a\over z-z_0}$. The function $g(z):=f(z)-{a\over z-z_0}$ has a removable singularity at $z_0$ and is analytic otherwise in $D_R$, which implies that $g$ is in fact analytic in $D_R$. So $g$ has a Taylor expansion $g(z)=\sum_{n=0}^\infty b_nz^n$ which is convergent for $|z|<R$, in particular at $z:={1\over2}(1+R)$. Therefore one necessarily has $$|b_n|\leq\Bigl({2\over1+R}\Bigr)^n\qquad(n>n_0)\ ;$$
which implies $\lim_{n\to\infty} b_n=0$. On the other hand,
$${a\over z-z_0}=-{a/z_0 \over 1-(z/z_0)}=-{a\over z_0}\sum_{n=0}^\infty\Bigl({z\over z_0}\Bigr)^n\ .$$
Now $f(z)={a\over z-z_0}+g(z)$. Therefore the Taylor coefficients $a_n$ of $f$ are given by
$$a_n=-{a\over z_0^{n+1}} + b_n\ ,$$
from which we get
$${a_n\over a_{n+1}}={-a/z_0^{n+1} +b_n\over -a/z_0^{n+2}+b_{n+1}}=z_0{a-b_n z_0^{n+1} \over a – b_{n+1}z_0^{n+2}}\to z_0\qquad(n\to\infty)\ .$$

Consider $\displaystyle f(z) = \frac{g(z)}{z-z_0}$. Since $f$ has a simple pole at $z_0$, $g$ is holomorphic on the disc.

Now, let $f = \sum a_nz^n, g=\sum b_nz^n$. Then, $\sum b_nz^n = (z-z_0)\sum a_nz^n$. Comparing the coefficients, you get $b_{n+1} = a_n – z_0a_{n+1}$.

As $n \to \infty$, $b_n \to 0$ while $a_n$ will not tend to $0$.