# Convergence of a power series at a point

Suppose a power series $\sum a_n z^n$ with $a_n,z \in \mathbb{C}$ converges for some single point $z_0$ in the complex plane.

Can we then say that it converges on any disk with radius $r\leq |z_0|$? I think yes since the power series has a radius of convergence, say $R$, where the series converges for $|z|<R$ and diverges for$|z|>R$. So, since the series converges at a single point, R must be 'big enough' to include this point.

Am I correct in my thinking?

• it converges for any $|z| < R$ but it doesn't have to converge at any other point on $|z| = R$. In general, a power series converges on $|z| < r$ and diverges for $|z| > r$ for some $r$ (possibly $r = \infty$ or $r = 0$), but everything is possible on $|z| = r$ (converges nowhere, everywhere, only at some points) – reuns Aug 3 '16 at 12:13
• What would be an example as in your first sentence? – zhw. Aug 4 '16 at 17:35

You can affirm that converges in a disk with $r<|z_0|$. Not equal. For example, $\sum \frac{1}{n}z^n$ converges for all $z\in \{z:|z|\leq 1\}-\{1\}$, but with $z=1$ diverges.
Then, if $z_0=i$, don't have convergency in the disk with $r=1$
• Does the power series converge for $z=i?$ – fosho Aug 3 '16 at 7:08
• @Dman You say $\sum(-1)^n\frac1n$ diverges? Hmm... – Did Aug 3 '16 at 7:09
• @vvnitram Your example and answer is correct, and thus +1 . Yet, if instead $\;i\;$ you'd choose $\;z_0=-1\;$ perhaps things could have gone smoother, as the alternating Harmonic series is the standard (or almost) example of (convergent, of course) Leibniz series. – DonAntonio Aug 3 '16 at 7:13