Let

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

Proposition 4

There exists ${\displaystyle r}$, ${\displaystyle 0\leqslant r\leqslant \infty }$, such that Radius of convergence is absolutely convergent if ${\displaystyle |z| and divergent if ${\displaystyle |z|>r}$. The convergence is moreover on any smaller closed disc ${\displaystyle K=\{z:|z|\leqslant r_{1}\}}$, (${\displaystyle r_{1}).

Proof

For any ${\displaystyle s\geqslant 0}$ let

${\displaystyle M(s)=\sup\{|a_{n}|s^{n}:n=0,1,2,\ldots \}}$
where we allow the possibility ${\displaystyle M(s)=\infty }$ if the sequence ${\displaystyle (|a_{n}|s^{n})}$ is unbounded. We note that if ${\displaystyle s, then ${\displaystyle M(s)\leqslant M(t)}$ (where this may have to be interpreted as ${\displaystyle \infty \leqslant \infty }$). To show this, suppose for example that ${\displaystyle M(t)<\infty }$. Then
${\displaystyle |a_{n}|s^{n}=|a_{n}|t^{n}(s/t)^{n}\leqslant M(t)(s/t)^{n}\leqslant M(t).}$
Thus every element of the sequence ${\displaystyle (|a_{n}|s^{n})}$ is ${\displaystyle \leqslant M(t)}$ and so the same is true of their sup ${\displaystyle M(s)}$. A similar argument shows that if ${\displaystyle M(s)=\infty }$ and ${\displaystyle s, then also ${\displaystyle M(t)=\infty }$.

Since ${\displaystyle M(0)=|a_{0}|<\infty }$, it makes sense to ask for the largest ${\displaystyle s}$ with ${\displaystyle M(s)<\infty }$. In other words, define

${\displaystyle r=\sup\{s:M(s)<+\infty \}.}$
The claim is that ${\displaystyle r}$ has the required properties of the Proposition. (Note that ${\displaystyle M(r)}$ can be either finite or ${\displaystyle +\infty }$, depending on the particular power series.)

If ${\displaystyle |z|, then there exists ${\displaystyle r_{1}}$ with ${\displaystyle |z|<\rho . We have seen that ${\displaystyle M(\rho )}$ is finite, and so

${\displaystyle |a_{n}||z^{n}|=|a_{n}|\rho ^{n}(|z|/\rho )^{n}\leqslant M(\rho )(|z|/\rho )^{n}.}$
The absolute convergence follows by comparision with the geometric series ${\displaystyle M(\rho )\sum (|z|/\rho )^{n}}$ which is convergent because ${\displaystyle |z|<\rho }$.

If ${\displaystyle |z|>r}$, then ${\displaystyle M(|z|)=+\infty }$ so that the sequence ${\displaystyle (|a_{n}||z|^{n})}$ is unbounded. Thus the terms in Radius of convergence don't tend to zero and the partial sums cannot converge.

Finally, let ${\displaystyle K=\{|z|\leqslant r_{1}\}}$ be a closed subdisc. Let ${\displaystyle \rho }$ be chosen so that ${\displaystyle r_{1}<\rho . Then we can run the above argument uniformly for all ${\displaystyle |z| (even ${\displaystyle \leqslant r_{1}}$),

${\displaystyle |a_{n}||z^{n}|=|a_{n}|\rho ^{n}(|z|/\rho )^{n}\leqslant M(\rho )(|z|/\rho )^{n}\leqslant M(\rho )(r_{1}/\rho )^{n}}$
if ${\displaystyle |z|\leqslant r_{1}}$. Uniform convergence on ${\displaystyle K}$ is equivalent to the following version of Cauchy's criterion:

Given ${\displaystyle \varepsilon >0}$, there exists ${\displaystyle N}$ (depending upon ${\displaystyle \varepsilon )}$ such that if ${\displaystyle m>n>N}$, then ${\displaystyle |\sum _{j=n}^{m}a_{j}z^{j}|<\varepsilon }$ for all ${\displaystyle z\in K}$.

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

${\displaystyle \left|\sum _{j=n}^{m}a_{j}z^{j}\right|\leqslant \sum _{j=n}^{m}|a_{j}||z|^{j}\leqslant M(\rho )\sum _{j=n}^{m}(r_{1}/\rho )^{j}\leqslant {\frac {(r_{1}/\rho )^{n}}{1-(r_{1}/\rho )}}}$
and this tends to zero as ${\displaystyle n\to \infty }$ if ${\displaystyle r_{1}<\rho }$. (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.)

Remark 5

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

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

Remark 6

Consider the three series

{\displaystyle {\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 ${\displaystyle r=1}$ (exercise). In the case of Radius of convergence and Radius of convergence ${\displaystyle M(1)}$, defined in Radius of convergence , is finite, in fact equal to ${\displaystyle 1}$. In the case of Radius of convergence , ${\displaystyle M(1)=+\infty }$.

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

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

Remark 7

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