Suppose the radius of convergence of $\sum_n a_n x^n$ is $r$ ($r$ is a positive number). Prove that the radius of convergence of $\sum_n a_n^2 x^n$ is $r^2.$
I've tried to use Cauchy–Hadamard theorem, but I think it's illegal to use arithmetic of limits when we talk about limsup.
Thank you.
You can use the ratio test. The ratio test says that if the limit as n goes to infinity of the absolute value of the ratio of successive terms in a series is less than 1, then the series converges. If greater than 1, then it does not.
The radius of convergence of the first series is r, so this means that the first series converges for all $x$ with $|x|<r$. In terms of the ratio test this means that for any $x$ with $|x|<r$ and no x with $|x|>r$ we have: $$ \lim_{n \to \infty} \left| \frac{a_{n+1}x^{n+1}}{a_n x^n} \right| < 1\\ \lim_{n\to \infty} \left| \frac{a_{n+1}x}{a_n} \right| <1\\ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| |x| <1\\ |x| \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| <1 $$
Since this is true for all $x$ with $|x|<r$ and no $x$ with $|x|>r$, it must be that $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{r}$.
Now consider the ratio test for the second sequence: $$ \lim_{n \to \infty} \left| \frac{(a_{n+1})^2x^{n+1}}{(a_n)^2 x^n} \right| \\ = \lim_{n \to \infty} \left| \frac{(a_{n+1})^2x}{(a_n)^2 } \right|\\ = \lim_{n \to \infty} \left| \frac{(a_{n+1})^2}{(a_n)^2} \right||x|\\ = |x| \lim_{n \to \infty} \left| \frac{(a_{n+1})^2}{(a_n)^2} \right|\\ = |x| \left( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\right)^2\\ = |x| \left(\frac{1}{r}\right)^2\\ = \frac{|x|}{r^2} $$ The series will converge if this is more than 1, and diverge if it's less than 1. It will be less than 1 when $|x| < r^2$ and more than 1 when $|x| > r^2$. So the radius of convergence of the second series must be $r^2$.