# Convergence of Taylor Series

My professor made this claim about Taylor Series convergence in my Complex Variables class and I am still not entirely convinced (he said it’s explained in the textbook and textbook states, “we will soon see that it is impossible for the series to converge to $f(z)$ outside this circle” but they don’t actually show that).

Basically the problem I’m having is this:

Let $f(z)$ be expanded in a Taylor series about $z_0$. Let $a$ be the distance from $z_0$ to the nearest singular point of $f(z)$. The claim is that the series converges to $f(z)$ everywhere in the circle $|z-z_0|=a$ and fails to converge to $f(z)$ outside the circle (while nothing can be determined about the points on the circle).

From what I’ve read in my book, I’m entirely convinced of the first claim that the series converges to $f(z)$ in the circle. It’s the other claim that it fails to converge to $f(z)$ outside the circle that is bothering me.

Here is what I know:

I know that the circle for which the Taylor series is everywhere convergent to $f(z)$ is not necessarily the same as the circle throughout which the series converges in general.

I know that $|z-z_0|=a$ is the largest circle within which the series converges to $f(z)$. All this says however is that we can’t construct a larger circle for which the series converges to $f(z)$; it still leaves open the possibility that the series could converge to $f(z)$ outside this circle.

I also have this one theorem we proved that states that if a power series centered at $z_0$ converges at the point $z_1$, then the series converges uniformly to an analytic function for all points in the disk $|z-z_0|<|z_1-z_0|$.

Now, the argument my professor made is that suppose that at the point $z=z_1$ the series converged to $f(z)$ where $z_1$ is outside the circle $|z-z_0|=a$. Then by the theorem above, the series would converge to an analytic function inside the circle $|z-z_0|<|z_1-z_0|$. (At this point he stated we had a contradiction and ended our conversation). This would seem like we are onto some kind of contradiction since $f(z)$ is clearly not analytic inside this circle but there is nothing that prevents the series from converging to something other than $f(z)$ outside the circle $|z-z_0|=a$.

I feel like I’m really missing something here or that there is some hole in my understanding. Any help is appreciated!

#### Solutions Collecting From Web of "Convergence of Taylor Series"

If a power series centered at $a$ converges absolutely at a point $w$, then it also converges at all points $|z-a|<|w-a|$.

If a power series converges absolutely then it converges to a holomorphic function $g$.

Finally, let $s$ be a singular point. The function $f$ and the function $g$ to which the power series converges coincide in the disc $|z-a|<|s-a|$. Therefore $g$ is an holomorphic extension of $f$ to a neighborhood of $s$. This contradics the definition of singular point.

Perhaps your book is referring to the possibility of the series converging on the boundary of the disc. It is true that the series may converge to $f$ on other points outside the open disc. But all these points at which it converges to $f$ must be on the boundary of the disc of convergence.

The main claim here is that any power series $\sum_{n = 0}^\infty a_n z^n$ converges everywhere inside an open disc and nowhere outside its closure. The proof is the root test, where since $\sqrt[n]{\lvert a_n z^n \rvert} = \lvert a_n \rvert^{1/n} \, \lvert z \rvert$, the test produces a condition $\lvert z \rvert < 1/L$, where $L$ is the limiting value of $\lvert a_n\rvert^{1/n}$. That is, the series converges inside the circle of this radius and diverges outside.

An even higher-level way of looking at it is that a power series is comparable, in the limit, to a geometric series, and geometric series have a radius of convergence. So the claim follows from the limit comparison test.

As for why the series can’t converge to $f$ in a larger disc containing a singular point of $f$: this actually depends on the definition of a singular point. Literally, if this happens, then the series itself defines a holomorphic function that agrees with $f$ around its singular point and also is defined at this point, where $f$ was said not to be holomorphic. As I learned it, that’s the definition of a “removable singularity” of $f$. The theorem as stated should require that the series converge up to the nearest non-removable singularity for this reason; note that your definition of singular point does not exclude removable ones (for example, one could simply redefine $f$ at one point to make it discontinuous, and thus singular, but this is still removable and doesn’t affect the power series centered anywhere else).