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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 $(\Omega)$.

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

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