Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $(\Omega)$.

share|cite|improve this question
up vote 4 down vote accepted

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 $\Omega=\{z \in \mathbb{C}:\ |z|\le 1\}$.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.