Let

$\sum _{n=0}^{\infty }a_{n}z^{n}$ be a power series centred at $z=0$ . Here the $a_{n}$ , the of the power series are fixed complex numbers.

Proposition (4):

There exists $r$ , $0\leqslant r\leqslant \infty$ , such that Radius of convergence is absolutely convergent if $|z| and divergent if $|z|>r$ . The convergence is moreover on any smaller closed disc

$K=\{z:|z|\leqslant r_{1}\}$ , ($r_{1} ).

Proof:

For any $s\geqslant 0$ let

$M(s)=\sup\{|a_{n}|s^{n}:n=0,1,2,\ldots \}$ where we allow the possibility $M(s)=\infty$ if the sequence $(|a_{n}|s^{n})$ is unbounded. We note that if $s , then $M(s)\leqslant M(t)$ (where this may have to be interpreted as $\infty \leqslant \infty$ ). To show this, suppose for example that $M(t)<\infty$ . Then
$|a_{n}|s^{n}=|a_{n}|t^{n}(s/t)^{n}\leqslant M(t)(s/t)^{n}\leqslant M(t).$ Thus every element of the sequence $(|a_{n}|s^{n})$ is $\leqslant M(t)$ and so the same is true of their sup $M(s)$ . A similar argument shows that if $M(s)=\infty$ and $s , then also $M(t)=\infty$ . Since $M(0)=|a_{0}|<\infty$ , it makes sense to ask for the largest $s$ with $M(s)<\infty$ . In other words, define
$r=\sup\{s:M(s)<+\infty \}.$ The claim is that $r$ has the required properties of the Proposition. (Note that $M(r)$ can be either finite or $+\infty$ , depending on the particular power series.) If $|z| , then there exists $r_{1}$ with $|z|<\rho . We have seen that $M(\rho )$ is finite, and so
$|a_{n}||z^{n}|=|a_{n}|\rho ^{n}(|z|/\rho )^{n}\leqslant M(\rho )(|z|/\rho )^{n}.$ The absolute convergence follows by comparision with the geometric series $M(\rho )\sum (|z|/\rho )^{n}$ which is convergent because $|z|<\rho$ . If $|z|>r$ , then $M(|z|)=+\infty$ so that the sequence $(|a_{n}||z|^{n})$ is unbounded. Thus the terms in Radius of convergence don't tend to zero and the partial sums cannot converge. Finally, let $K=\{|z|\leqslant r_{1}\}$ be a closed subdisc. Let $\rho$ be chosen so that $r_{1}<\rho . Then we can run the above argument uniformly for all $|z| (even $\leqslant r_{1}$ ),
$|a_{n}||z^{n}|=|a_{n}|\rho ^{n}(|z|/\rho )^{n}\leqslant M(\rho )(|z|/\rho )^{n}\leqslant M(\rho )(r_{1}/\rho )^{n}$ if $|z|\leqslant r_{1}$ . Uniform convergence on $K$ is equivalent to the following version of Cauchy's criterion:
Given $\varepsilon >0$ , there exists $N$ (depending upon $\varepsilon )$ such that if $m>n>N$ , then $|\sum _{j=n}^{m}a_{j}z^{j}|<\varepsilon$ for all $z\in K$ .

This is satisfied in our case for $K=\{|z|\leqslant r_{1}\}$ because

$\left|\sum _{j=n}^{m}a_{j}z^{j}\right|\leqslant \sum _{j=n}^{m}|a_{j}||z|^{j}\leqslant M(\rho )\sum _{j=n}^{m}(r_{1}/\rho )^{j}\leqslant {\frac {(r_{1}/\rho )^{n}}{1-(r_{1}/\rho )}}$ and this tends to zero as $n\to \infty$ if $r_{1}<\rho$ . (In the last inequality in Radius of convergence we have bounded a finite geometric progression by its sum to infinity. This is a device that is often

handy.)

$\Box$ Remark:

It follows from the definition that if Radius of convergence has radius of convergence $r$ , then so do the related series

$\sum _{n=N}^{\infty }a_{n}z^{n}{\mbox{ and }}\sum _{n=N}^{\infty }a_{n}z^{n+k}$ where $N\geqslant 0$ is an integer, $k$ is an integer, and we assume $N+k\geqslant 0$ if $k$ is negative.

Thus the radius of convergence is unaffected by throwing away any finite number of terms in the sum, or by multiplying by a fixed power of $z$ .

Remark:

Consider the three series

{\begin{aligned}\sum _{n=0}^{\infty }z^{n};&&\\\sum _{n=0}^{\infty }nz^{n};&&\\\sum _{n=0}^{\infty }{\frac {z^{n}}{n^{2}}}.\end{aligned}} In each case the radius of convergence $r=1$ (exercise). In the case of Radius of convergence and Radius of convergence $M(1)$ , defined in Radius of convergence , is finite, in fact equal to $1$ . In the case of Radius of convergence , $M(1)=+\infty$ .

We note also that these behave differently at the boundary of the radius of convergence, when $|z|=1$ . In cases (a) and (b), the series diverge for all $z$ with $|z|=1$ . (The terms don't even go to zero.) In case (c), the series converges for all values of $z$ with $|z|=1$ .

Thus you cannot say anything about a power series at its radius of convergence: the behaviour can be arbitrarily wild.

Remark:

I wrote the section on power series before the section on uniform convergence. The arguments may seem quite repetitive. In fact, with the $M$ -test in hand, the discussion of power series could have been streamlined by using the $M$ -test with $M_{n}=M(\rho )(r_{1}/\rho )^{n}$ for the last part of the proof of Proposition~ Radius of convergence .