Radius of convergence for $a_n=\frac{1}{n^2}$. I am asked to find the radius of convergence for the power series
$$\sum_{n=0}^{\infty} a_nz^n$$ with $a_n=\frac{1}{n^2}$.
I did this very complicated with $e^{ln(\frac{1}{n^2})^{1/n}}$. is there an easier way of doing this? maybe the ratio test?
 A: In a standard calculus course, the radius of convergence is defined as: $r=\displaystyle \lim_{n \to \infty} \dfrac{|a_n|}{|a_{n+1}|}= \displaystyle \lim_{n \to \infty} \dfrac{(n+1)^2}{n^2}=1$. You can prove it yourself that this is the case.
A: Here is a (extended) version from WangYeFei's answer.
According to the ratio test, the power series converges whenever $|z|$ satisfies the following relation:
\begin{align*}
\lim_{n\to\infty}\left|\frac{a_{n+1}z^{n+1}}{a_{n}z^{n}}\right| & = \lim_{n\to\infty}\left|\frac{z^{n+1}}{(n+1)^{2}}\times\frac{n^{2}}{z^{n}}\right|\\\\
& = \lim_{n\to\infty}\left|z\times\left(\frac{n}{n + 1}\right)^{2}\right|\\\\
& = |z| < 1
\end{align*}
So the answer to your question is given by $r = 1$.
Hopefully this helps!
A: If $|z|>1$ let $|z|=1+s$ with $s>0.$ Then for $n\ge 2$ we have $$|a_nz^n|=n^{-2}\sum_{j=0}^n\binom n j s^j\ge$$ $$\ge n^{-2}\binom n 2 s^2=$$ $$=\frac {s^2}{2}(1-\frac 1 n)\ge$$ $$\ge \frac {s^2}{4}$$ so the terms of the series do not $\to 0$ so the series is not summable.
If $|z|<1$ the series converges by comparison with the absolutely convergent series $\sum_n|z|^n.$
