# How to determine the radius of convergence if the Taylor series cannot be written in a neat way?

I am trying to evaluate the radius of convergence of Taylor series centered at zero of function $$f(z)=\frac{\sin(3z)}{\sin(z+\pi/6)}$$ I guess the answer should be $\pi/6$ because the function will not be bounded if $x$ approaches $\pi/6$. And it is easy to show that $f$ is convergent for all $z$ with norm less than $\pi/6$ because the denominator will not reach zero.

However, outside the circle with radius $\pi/6$ there do exist points that makes $f$ converge. So I am confused whether I get the right answer.

Usually for a simple power series, if we determine the radius of convergence we will have the series diverge for all $z$ outside the circle. So I am really not sure what is the situation here.

• Yes, up to the first singularity. Feb 12, 2015 at 4:39
• Thanks for the remind. I have edited it.
– user194201
Feb 12, 2015 at 5:01

The power series will diverge outside the disk of radius $\frac\pi6$ centered at $0$. That doesn't mean the function isn't defined there.
"Usually for a simple power series..." This is the usual. Consider the function $f(z)=\frac1{1-z}$. Its power series centered at $0$ is $\sum\limits_{n=0}^\infty z^n$, with radius of convergence $1$, even though $f(z)$ is defined when $|z|>1$. Outside the disk of convergence, these functions have different power series with different centers.
• No, we can say that the power series definitely does not converge at any point outside of the circle, regardless of whether or not the original function is defined there. E.g., $\frac{1}{1-z}$ is defined when $z=2$, but $1+2+2^2+2^3+2^4+\cdots$ diverges. If you review properties of the radius of convergence, you will find it is characterized by the fact that the series converges everywhere within the circle, nowhere outside, regardless of how nice the function might be elsewhere. Feb 12, 2015 at 5:01