I'm currently trying to answer the following complex analysis question: "give an example of a power series $\sum\limits_{n=0}^{\infty}a_nz^n$ with radius of convergence $2$ which converges at some $z_0 \in \{z: |z|=2\}$ and diverges at some $z_1 \in \{z: |z|=2\}$".
This is what I currently have:
"Consider the following series: $\sum\limits_{n=0}^{\infty}(\frac{1}{2})^nz^n$.
Then we have that the radius of convergence is:
$R=\lim_{n\to\infty}|\frac{a_n}{a_{n+1}}|
=\lim_{n\to\infty}|\frac{(1/2)^n}{(1/2)^{n+1}}|
=|\frac{1}{1/2}|$
$=|2|$
$=2$.
Let $z_1=2$.
Then we have $\sum\limits_{n=0}^{\infty}(\frac{1}{2})^n(2)^n$
$=\sum\limits_{n=0}^{\infty}(1/2\cdot2)^n$
$=\sum\limits_{n=0}^{\infty}(1)^n$
$=\sum\limits_{n=0}^{\infty}1$, which diverges."
My question is how to find a $z_0 \in \{z: |z|=2\}$ where my series converges. Any help would be greatly appreciated!
Take $a_0 = 0$ and $a_n = \frac{1}{n2^n}$ for $n > 0$. Then $R = 2$ and $\sum_{n=0}^\infty a_n 2^n = \sum_{n=1}^\infty \frac{1}{n}$ diverges, whereas $\sum_{n=0}^\infty a_n (-2)^n = \sum_{n=1}^\infty (-1)^n\frac{1}{n}$ converges.