Let $$\sum_{n=0}^{\infty}a_nx^n,\quad \sum_{n=0}^{\infty}b_nx^n$$ be two power series with radius of convergence $R_1$ and $R_2$ satisfying $$0<R_1<R_2<+\infty,\ a_n\neq0,n\in\mathbb{N}.$$ Can we have the following result: $$\lim_{n\to\infty}\frac{b_n}{a_n}=0.$$
What I have try: If $$\frac{1}{R_1}=\lim_{n\to\infty}\sqrt[n]{|a_n|}\ \mbox{and}\ \frac{1}{R_2}=\lim_{n\to\infty}\sqrt[n]{|n_n|},$$ it is easy to show the above result. But,in general, $$\frac{1}{R_1}=\limsup_{n\to\infty}\sqrt[n]{|a_n|}\ \mbox{and}\ \frac{1}{R_2}=\limsup_{n\to\infty}\sqrt[n]{|b_n|}.$$
Other method: If we choose $r\in(R_1,R_2)$, then $$\sum_{n=0}^{\infty}b_nr^n$$ is convergent, and $$\sum_{n=0}^{\infty}a_nr^n$$ is divergent, and $$b_nr^n=\frac{b_n}{a_n}\cdot a_nr^n\to0.$$ If we know $|a_nr^n|\to \infty$, then we will get $$\lim_{n\to\infty}\frac{b_n}{a_n}=0.$$
Any hlep and hints will welcome or counterexample can be provided!
A counterexample:
Then $R_1 = 1 < 2 = R_2$, and $b_{2n}/a_{2n} \to \infty$.
Similarly one can construct examples such that $b_n/a_n$ has arbitrary prescribed subsequential limits.
One can only conclude that $\liminf_{n \to \infty} b_n/a_n = 0$. In particular, if the limit exists, it must be zero.