Finding radii of convergence of simple power series using the ratio test

Question

I am having a hard time finding the radii of convergence using the ratio test.
Let $\sum_{n=1}^{\infty}a_nx^n$ be a power series with radius of convergence $R$. I need to find the radii of convergence of $\sum_{n=1}^{\infty}\frac{1}{a_n}x^n$ and $\sum_{n=1}^{\infty}a_nx^{2n}$.
My try: Using the ratio test, I know that $\lim|\frac{a_{n+1}}{a_n}x|<1$ for all $x\in(-R,R)$. I have to find all $x$ such that $\lim|\frac{a_{n}}{a_{n+1}}x|<1$ and all $x$ such that $\lim|\frac{a_{n+1}}{a_{n}}x^{n+2}|<1$. Can anyone give me some hint how to get started?

Answer 1

I will assume that each $a_n$ is different from $0$.

You cannot know the radius of convergence of $\sum_{n=1}^\infty\frac1{a_n}x^n$ from the redius of convergence of $\sum_{n=1}^\infty a_nx^n$. Take $a>1$ and take$$a_n=\begin{cases}a^n&\text{ if $n$ is odd}\\1&\text{ if $n$ is even.}\end{cases}$$Then the radius of convergence of $\sum_{n=1}^\infty a_nx^n$ is $\frac1a$, but the radius of convergence of $\sum_{n=1}^\infty\frac1{a_n}x^n$ is always $1$.

On the other hand, if the radius of convergence of $\sum_{n=1}^\infty a_nx^n$ is $R$, then the radius of convergence of $\sum_{n=1}^\infty a_nx^{2n}$ is $\sqrt R$, since $\sum_{n=1}^\infty a_nx^{2n}=\sum_{n=1}^\infty a_n\left(x^2\right)^n$, which converges when $x^2<R$ and diverges when $x^2>R$.