The question is #$14$ from Chapter $2$ in Stein and Shakarchi's text Complex Analysis:

Suppose that $f$ is holomorphic in an open set containing the closed unit disc, except for a pole at $z_0$ on the unit circle. Show that if $$\sum_{n=0}^\infty a_nz^n$$ denotes the power series expansion $f$ in the open unit disc, then $$\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=z_0.$$

I've shown that we can take $z_0=1$ without a loss of generality, but I'm having trouble showing the proof otherwise. One of the problems I'm having is because we aren't told the definition of a pole except that it is a place where the function isn't holomorphic. Disregarding this fact, the other problem I'm running into is that I don't know the order of the pole.

Making some additional assumptions, including that the pole is simple so we can write $F(z)=(z-1)f(z)$ as a holomorphic function, we see that $$F(z)=-a_0+z(a_0-a_1)+z^2(a_1-a_2)+\cdots$$ This almost gets me to the end with these added assumptions, but I don't think it's quite enough (why do we know some of the $a_i$'s aren't $0$, for example).

On another note, if we know $\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ exists, then it is easy to see that $\lim_{n\to\infty}\frac{|a_n|}{|a_{n+1}|}=1=|z_0|$; I, however, do not see why the limit must exist.

Are there any hints that someone can provide? Even a solution would be nice, especially if one can avoid making any assumptions about what a pole is or is not.

EDIT: So there isn't any confusion, I know the definition of a pole and I'm inclined to believe that the problem, as stated, necessarily has a pole at $z_0$. The problem is that the exercise is in Chapter $2$, and poles are introduced in Chapter $3$.

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    $\begingroup$ I don't have Stein and Shakarchi on me, but it's certainly not correct that a pole is the same thing as a place where the function is not holomorphic. Being a pole is a much more restrictive condition: it means that $(z-1)^m f$ is holomorphic at $z = 1$ for some positive integer $m$. $\endgroup$ Jun 14, 2013 at 23:32
  • $\begingroup$ I didn't actually sit down and work the whole thing out, but I think the approach you sketched above is at least approximately the right way to attack the problem. $\endgroup$ Jun 14, 2013 at 23:33
  • $\begingroup$ @Clayton There's more involved with being a pole than just failing to be holomorphic. For example $e^{1/z}$ fails to be holomorphic at $0$ but does not have a pole there. You will necessarily need this information to solve the problem. $\endgroup$
    – Potato
    Jun 14, 2013 at 23:52
  • $\begingroup$ @Clayton I'm fairly sure this is false if 'pole' is replaced by 'essential singularity,' so as I said, you will require the characterization of a pole Daniel mentioned to solve this. This means it cannot be solved without referring to material in chapter 3. (These books are good, but not the most well-edited things in the world. It does not surprise me there's a minor glitch like this.) $\endgroup$
    – Potato
    Jun 15, 2013 at 0:00
  • $\begingroup$ @Clayton So if the problem is really finding a counterexample in the case of an essential singularity, you should edit the question to say that. As it's written right not, it just looks like you're confused about what a pole is. $\endgroup$
    – Potato
    Jun 15, 2013 at 0:11

4 Answers 4


I will assume we can write

$$f(z) = \frac{c}{z_0-z} + \sum_{n=0}^{\infty} b_n z^n$$

for some value of $c \ne 0$, and $\lim_{n \to \infty} b_n = 0$. Then

$$f(z) = \sum_{n=0}^{\infty} a_n z^n$$


$$a_n = b_n + \frac{c}{z_0^{n+1}}$$


$$\begin{align}\lim_{n \to \infty} \frac{a_n}{a_{n+1}} &= \lim_{n \to \infty} \frac{\displaystyle b_n + \frac{c}{z_0^{n+1}}}{\displaystyle b_{n+1}+ \frac{c}{z_0^{n+2}}}\\ &= \lim_{n \to \infty} \frac{\displaystyle \frac{c}{z_0^{n+1}}}{\displaystyle \frac{c}{z_0^{n+2}}}\\ &= z_0\end{align}$$

as was to be shown. Note that the second step above is valid because $z_0$ is on the unit circle.

For a nonsimple pole, we may write

$$f(z) = \frac{c}{(z_0-z)^m} + \sum_{n=0}^{\infty} b_n z^n$$

for $m \in \mathbb{N}$. It might be known that

$$(1-w)^{-m} = \sum_{n=0}^{\infty} \binom{m-1+n}{m-1} w^n$$


$$\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \frac{n+1}{n+m} z_0$$


@TCL observed that we can simply require that $b_n z_0^n$ goes to zero as $n \to \infty$. Then for a simple pole

$$\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \frac{\displaystyle b_n z_0^n + \frac{c}{z_0}}{\displaystyle b_{n+1} z_0^n + \frac{c}{z_0^2}}$$

which you can see goes to $z_0$.

  • $\begingroup$ @Potato: thanks! $\endgroup$
    – Ron Gordon
    Jun 14, 2013 at 23:52
  • $\begingroup$ I thought $a_n = b_n + \frac{c}{z_0^{n+1}}$? Why is there a $(-1)^n$? $\endgroup$
    – TCL
    Jun 15, 2013 at 0:04
  • $\begingroup$ @RonGordon: The procedure of taking the limit of $b_n$ and not of the rest is suspect and probably wrong. $\endgroup$ Jun 15, 2013 at 0:04
  • $\begingroup$ @Clayton: good point, I mistakenly used the wrong expansion and saw abs value signs. Fixed. $\endgroup$
    – Ron Gordon
    Jun 15, 2013 at 0:05
  • $\begingroup$ @TCL: a mistake. Fixed. $\endgroup$
    – Ron Gordon
    Jun 15, 2013 at 0:06

This is just an attempt to give a rigorous proof of Ron Gordon's answer.

Suppose the order of the pole at $z_0$ is $m$. Then for some constants $c_0,\ldots,c_m, c_m\neq 0,$ $$g(z)=f(z)-\left(\frac{c_m}{(z_0-z)^m}+\cdots +\frac{c_1}{(z_0-z)}\right)$$ is analytic at $z_0$. Since $f(z)$ is analytic on an open set containing the unit disk except at $z_0$, we see that $g(z)$ is analytic on an (possibly different) open set containing the unit disk. So $$g(z)=\sum_{n=0}^\infty b_nz^n$$ has radius of convergence greater than 1, in particular the series converges at $z_0$ and $\lim_n b_nz_0^n=0$. So for $|z|<1$, $$f(z)=\sum_{n=0}^\infty a_nz^n=\sum_{n=0}^\infty b_nz^n+\left(\frac{c_m}{(z_0-z)^m}+\cdots +\frac{c_1}{(z_0-z)}\right)$$ For $1\le k\le m$, we have $$\frac{c_k}{(z_0-z)^k}=\sum_{n=0}^\infty \frac{(k)_nc_k}{n!z_0^{n+k}} z^n$$ where $(k)_n=k(k+1)\cdots(k+n-1)$ and $(k)_0=1$. It follows that $$a_n=b_n+\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{n+k}}$$ and \begin{align*}\frac{a_n}{a_{n+1}}&=\frac{\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{k}} +b_nz_0^n}{\sum_{k=1}^m \frac{(k)_{n+1}c_k}{(n+1)!z_0^{1+k}}+b_{n+1}z_0^n}\\ &= z_0 \frac{\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{k}} +b_nz_0^n}{\sum_{k=1}^m \frac{(k)_{n+1}c_k}{(n+1)!z_0^{k}}+b_{n+1}z_0^{n+1}} \end{align*} Now divide the numerator and denominator of the last expression by $$\frac{(m)_{n+1}}{(n+1)!}$$, which is equal to 1 when $m=1$ and approaches infinity as $n\to\infty$ if $m>1$. Furthermore, for $1\le k\le m$, $$\frac{(k)_n}{n!}\cdot \frac{(n+1)!}{(m)_{n+1}}=\frac{(k)_n (n+1)}{(m)_n (n+m)}$$ approaches 0 as $n\to\infty$ if $k<m$ and approaches 1 if $k=m$. It follows that the fraction $$\frac{\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{k}} +b_nz_0^n}{\sum_{k=1}^m \frac{(k)_{n+1}c_k}{(n+1)!z_0^{k}}+b_{n+1}z_0^{n+1}}$$ approaches 1 as $n\to\infty$, and the proof is complete.

EDIT. If $k<m$, $$\frac{(k)_n }{(m)_n }=\prod_{i=0}^{n-1} \frac{k+i}{m+i}=\prod_{i=0}^{n-1} \left(1-\frac{m-k}{m+i}\right)\to 0$$ because $\sum_{i=0}^\infty \frac{m-k}{m+i}=\infty$.


Suppose that $$ \begin{align} f(z)(z-z_0)^n &=(z-z_0)^n\sum_{k=0}^\infty a_kz^k\\ &=\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}z^jz_0^{n-j}\sum_{k=0}^\infty a_kz^k\\ &=\sum_{k=0}^\infty\left(\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}a_{k-j}z_0^{n-j}\right)z^k \end{align} $$ converges in a neighborhood of $z_0$. Thus, for some $r\lt1$, we have $$ \left|\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}a_{k-j}z_0^{n-j}\right|\le c\,r^k $$ That is, the the $n^{\text{th}}$ finite difference of the sequence $a_kz_0^k$ satisfies $$ \Delta^na_kz_0^k=O(r^k) $$ Inverting the finite difference operator yields that there is an $n-1$ degree polynomial $P$ so that $$ a_kz_0^k=P(k)+O(r^k) $$ Taking the limit of the ratio of terms yields $$ \lim_{k\to\infty}\frac{a_kz_0^k}{a_{k+1}z_0^{k+1}}=\lim_{k\to\infty}\frac{P(k)+O(r^k)}{P(k+1)+O(r^{k+1})}=1 $$ That is, $$ \lim_{k\to\infty}\frac{a_k}{a_{k+1}}=z_0 $$

Inverting Finite Difference Operators

If we define the finite difference operator as $$ \Delta a_k=a_k-a_{k-1} $$ then, as with indefinite integrals, when inverting $\Delta$, we need to include a constant: $$ \Delta^{-1}a_k=c+\sum_{j=1}^k a_j $$ Suppose that $\Delta a_k=P_n(k)+O(r^k)$, where $P_n$ is a degree $n$ polynomial and $0\le r\lt1$. Since the sum of a degree $n$ polynomial is a degree $n+1$ polynomial and the sum of $O(r^k)$ is $c+O(r^k)$, $a_k=P_{n+1}(k)+O(r^k)$.

Iterating, we get that if $\Delta^n a_k=O(r^k)$, then $a_k=P_{n-1}(k)+O(r^k)$.

  • 1
    $\begingroup$ I don't see why $ \left|\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}a_{k-j}z_0^{n-j}\right|\le c\,r^k $ $\endgroup$
    – TCL
    Jun 15, 2013 at 19:56
  • 1
    $\begingroup$ @TCL: The power series for an analytic function will converge out to the nearest pole. For example, the radius of convergence for the power series for $\frac1{1-z}$ is $1$ because of the pole at $z=1$. It is given that $(z-z_0)^nf(z)$ is holomorphic in an open set containing the closure of the unit disc for some $n$ ($z_0$ is not an essential singularity since it is a pole). That means that the radius of convergence of $$\sum_{k=0}^\infty\left(\sum_{j=0}^n(-1)^{n-j} \binom{n}{j}a_{k-j}z_0^{n-j}\right)z^k$$ is greater than $1$. $\endgroup$
    – robjohn
    Jun 15, 2013 at 22:12
  • $\begingroup$ Thanks.+1. Where can I find info about inverting finite difference operator? $\endgroup$
    – TCL
    Jun 16, 2013 at 1:57
  • 1
    $\begingroup$ @TCL: I have added a small section on inverting finite difference operators on the relevant functions. $\endgroup$
    – robjohn
    Jun 16, 2013 at 4:38
  • $\begingroup$ I think it should be $$f(z)(z-z_0)^n=\sum_{k=0}^\infty\left(\sum_{j=0}^k(-1)^{n-j}\binom{n}{j}a_{k-j}z_0^{n-j}\right)z^k,$$ after you commutate the double summation. $\endgroup$
    – Riemann
    May 20, 2021 at 8:00

We can do this by induction on $m$ where $m$ is the order of the pole.

For a simple pole assume we have proved the result.

Then suppose $m>1$, then $(z-z_0)^mf(z)=g(z)$ where $g(z)$ is holomorphic on a disk of radius bigger than $1$.

Consider $h_i(z):= (z-z_0)^{m-i}f(z)$ for $1\leq i \leq m$.
Each $h_i$ is holomorphic on $B(0,1)$ and has a pole of order $i$ at $1$.

$h_i(z)= \sum_{n=0}^{\infty} a_{in} z^n, |z|<1$

Since $h_m(z)=f(z)$ we would be done if we show $\lim_{n\to \infty} \frac{a_{in}}{a_{i(n+1)}}=z_0 \Rightarrow \lim_{n\to \infty} \frac{a_{(i+1)n}}{a_{(i+1)n}}=z_0$ Note: We use the induction hypothesis to get this for $h_1$.

To this end, we make the following observation.


Comparing coefficients we deduce


and hence the following

$a_{(i+1)n}z_0^{n+1}= -(a_{i0}+a_{i1}z_0 + a_{i2} z_0^2 + \ldots + a_{in}z_0^n)$

Now, $\lim_{n\to \infty} \frac{a_{(i+1)n}}{a_{(i+1)n}} = z_0 \lim_{n\to \infty} \frac{a_{(i+1)n}z_0^n}{a_{(i+1)n}z_0^{n+1}}$.

Thus it is enough to show that

$\lim_{n\to \infty} \frac{a_{(i+1)n}z_0^n}{a_{(i+1)n}z_0^{n+1}}= 1$

$\lim_{n\to \infty} \frac{a_{i0}+a_{i1}z_0 + a_{i2} z_0^2 + \ldots + a_{in}z_0^n}{a_{i0}+a_{i1}z_0 + a_{i2} z_0^2 + \ldots + a_{in}z_0^n + a_{i(n+1)}z_0^{n+1}}=1$

$\lim_{n \to \infty} \frac{S_n(z_0)}{S_{n+1}(z_0)}=1$

where $S_n(z)$ is the $n^{th}$ partial sum at $z$ of $h_i$.

But the above limit is same as $\lim_{n \to \infty} \lim_{z\to z_0} \frac{(z-z_0)^iS_n(z)}{(z-z_0)^iS_{n+1}(z)} $

Then since the inner term converges uniformly in $z$ (independent of $n$) we can interchange the limits, which gives

$ \lim_{z\to z_0} \lim_{n \to \infty} \frac{(z-z_0)^iS_n(z)}{(z-z_0)^iS_{n+1}(z)}=\lim_{z \to z_0} \frac{g(z)}{g(z)}=1$

Please let me know if anything is unclear.


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