# Why is the domain of convergence of a power series a perfect disk?

I've been going over power series in my Differential Equations class for approximating solutions, and one thing that's been fascinating me is the statement that there is a radius of convergence, around the point for which the series converges. Now, I understand that when a function has an asymptote, like $f(x)=\frac1x$, this is naturally going to "obstruct" convergence, and so the series has to stop converging for values too close to the asymptote. But why should this hinder convergence on the other side of the point?

Furthermore, this implies that when dealing with a complex power series, the region of convergence is a perfect disk. Why can't we have more irregular regions of convergence? What keeps convergence about the point perfectly symmetric?

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It's not quite a perfect disk; the boundary behaviour can be rather complex. But to see why convergence at $z = r$ implies convergence at $z = s$ whenever $|r| > |s|$, use a comparison test. – user61527 Jun 13 '14 at 1:57

The set of points where a power series converges is not necessarily a perfect disk - convergence at one endpoint does not imply convergence at the other endpoint. For a particular example, the series

$$\sum_{n = 0}^{\infty} x^n$$

has radius of convergence $1$ (as easily deduced from the ratio test) and diverges at both $x = \pm 1$. On the other hand, the series

$$\sum_{n = 1}^{\infty} \frac{x^n}{n}$$

again has radius of convergence $1$ (again, from the ratio test perhaps). However, this diverges at the right endpoint, but converges at the left endpoint of $-1$; here, it is the alternating harmonic series.

So the boundary behaviour can be quite complicated (and even moreso in the complex case): The set of points on the boundary of the disk where the series converges can be large or small depending on the particular series.

All that we can say, generally, is that absolute convergence (and, in fact, substantially weaker versions of convergence) at a point with $z = r$ implies absolute convergence at $z = s$ whenever $|r| \ge |s|$; this follows immediately from a comparison test between the two series. since $|s|^n \le |r|^n$ for all $n$ in this case. This is the obstruction that keeps a series from converging absolutely at a boundary point, unless it converges absolutely at all boundary points.

Perhaps you'd find it easier to think about it in a slightly opposite manner: Convergence at a point implies convergence at all points closer to $0$ - so having a radius of convergence that's "too large" would be an obstruction to having a pole of any sort.

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Actually, simple convergence at a point with $z=r$ implies absolute convergence at $z=s$. – lhf Jun 13 '14 at 2:04
@lhf Indeed it does - I didn't mention this since it takes a little bit more work to deal with the question of why the series converges at closer points. – user61527 Jun 13 '14 at 2:05
... and even more, boundedness of the terms at $z=r$ implies absolute convergence at $z=s$. – Robert Israel Jun 13 '14 at 2:05
@RobertIsrael Now that is really something. How does that even work? Would a proof be too much work? – silvascientist Jun 13 '14 at 2:44
It's very simple. If $|a_n| |r|^n \le B$ and $|s| < |r|$, then $|a_n| |s|^n \le B t^n$ where $t = |s|/|r| < 1$, and $\sum_n B t^n$ converges. – Robert Israel Jun 13 '14 at 5:38

The key result is that if a power series converges at $z=z_1$ then it converges absolutely for all $z$ with $|z|<|z_1|$, and here you get your disk. The proof uses that $a_n z_1^n \to 0$ and a comparison with the geometric series.

As others have mentioned, this results says nothing about convergence on the circle $|z|=|z_1|$.

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