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 $|z|<r$ and divergent if $|z|>r$. The
convergence is moreover on any smaller closed disc
$K=\{z:|z|\leqslant r_{1}\}$, ($r_{1}<r$).

*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

$s<t$, 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

$s<t$, 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 $|z|<r$, then there exists $r_{1}$ with $|z|<\rho <r$. 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 $r_{1}<\rho <r$. Then we can run the above argument uniformly
for all $|z|<r_{1}$ (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.)

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 $z$.

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.

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** .