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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!

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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.

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

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