# Does $R_1<R_2$ impliy $\frac{b_n}{a_n}\to 0$ for two power series?

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 $$0 Can we have the following result: $$\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!

## 1 Answer

A counterexample:

• $$a_n = 1$$ for odd $$n$$, $$a_n = 1/3^n$$ for even $$n$$.
• $$b_n = 1/2^n$$.

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.

• Nice answer! Could you give some extra conditions making the result is right? May 19, 2021 at 8:24
• @Riemann: A simple condition would be that the limit exists: In that case it must be zero. May 19, 2021 at 8:28
• I see, thank you very much! May 19, 2021 at 8:32