# A doubt about complex variable.

Find the radius of convergence of the power series:$$\sum_{n\geq0}\frac{1}{(n+3)^2}z^n$$ Describes the convergence domain $(\Omega)$ and study the convergence point on the border $(\Omega)$.

-
Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many would consider your post rude because it is a command ("Find..."), not a request for help, so please consider rewriting it. –  Zev Chonoles Feb 4 '13 at 1:05
Sorry :( of course, I am new but I don't want to disturb. Thank you for the advice ^^ –  Sophie Germain Feb 4 '13 at 1:18
Don't worry about it :) There are always norms in any new community that take some explanation, and some getting used to. –  Zev Chonoles Feb 4 '13 at 1:21

Since $$\lim_{n\to \infty}\Big|\frac{z^{n+1}}{(n+4)^2}\cdot\frac{(n+3)^2}{z^n}\Big|=\lim_{n\to \infty}\Big|\frac{n+3}{n+4}\Big|^2|z|=|z|,$$ therefore if $|z|<1$, then the series converges. For $|z|=1$ the series is absolutely convergent (and therefore convergent) because $$\sum_{n=0}^\infty\frac{|z|^n}{(n+3)^2}=\sum_{n=3}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}-1-\frac14.$$ Hence the series converges in $\Omega=\{z \in \mathbb{C}:\ |z|\le 1\}$.