# The Unapologetic Mathematician

## Uniform Convergence of Power Series

So, what’s so great right now about uniform convergence?

As we’ve said before, when we evaluate a power series we get a regular series at each point, which may or may not converge. If we restrict to those points where it converges, we get a function. That is the series of functions converges pointwise to a limiting function. What’s great is that for any compact set contained within the radius of convergence of the series, this convergence is uniform!

To be specific, take a power series $\sum\limits_{n=0}^\infty c_nz^n$ which converges for $|z|, and let $T$ be a compact subset of the disk of radius $R$. Now the function $|z|$ is a continuous, real-valued function on $T$, and the image of a compact space is compact, so $|z|$ takes some maximum value on $T$.

That is, there is some point $p$ so that for every point $z\in T$ we have $|z|\leq|p|. And thus we have $|c_nz^n|\leq|c_np^n|$ for all $z\in T$. Setting $M_n=|c_np^n|$, we invoke the Weierstrass M-test — the series $\sum\limits_{k=0}^\infty|c_np^n|$ converges because $p$ is within the disk of convergence, and thus evaluation at $p$ converges absolutely.

Now every point within the disk of convergence is separated by some compact set (closed disks are compact, so pick a radius small less than the distance from the point to the boundary of the disk of convergence), within which the convergence is uniform. Since each term is continuous, the uniform limit will also be continuous at the point in question. Thus inside the radius of convergence a power series evaluates to a continuous function.

This gives us our first hint as to what can block a power series. As an explicit example, consider the geometric series $\sum\limits_{n=0}^\infty z^n$, which converges for $|z|<1$ to the function $\frac{1}{1-z}$. This function is clearly discontinuous at $z=1$, and so the power series can’t converge in any disk containing that point, since if it did it would have to be continuous there. And indeed, we can calculate the radius of convergence to be exactly ${1}$.

It’s important to note something in this example. For $|z|<1$, we have $\frac{1}{1-z}=\sum\limits_{k=0}^\infty z^n$, but these two functions are definitely not equal outside that region. Indeed, at $z=-2$ the function clearly has the value $\frac{1}{3}$, while the geometric series diverges wildly. The equality only holds within the radius of convergence.