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Suppose that $\lim\limits_{n\rightarrow \infty} |\frac{a_{n+1}}{a_n}| = \frac{1}{\alpha}$, $\lim\limits_{n\rightarrow \infty} |\frac{b_{n+1}}{b_n}| = \frac{1}{\beta}$ and $\alpha > \beta$. Does it implies that $\lim\limits_{n\rightarrow \infty} |\frac{a_n}{b_n}| = 0 ?$

I think it is correct because the condition means that increasing rate of $b_n$ greater than increasing rate of $a_n$. Then $\lim\limits_{n\rightarrow \infty} |\frac{a_n}{b_n}| = 0 $ no matter what initial value $a_0$ and $b_0$ are given.

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We show that $|b_n/a_n|\to 0$ if and only if $\alpha<\beta$.

If. Indeed, to be meaningful, it means that $\alpha,\beta>0$. Fix $\varepsilon>0$, hence there exists $n_0=n_0(\varepsilon)>0$ such that $$ |a_{n+1}|\ge \left(\frac{1}{\alpha}-\varepsilon\right)|a_n| \,\,\text{ and }\,\,|b_{n+1}|\le \left(\frac{1}{\beta}+\varepsilon\right)|b_n| $$ for all $n\ge n_0$. This implies $$ \left|\frac{b_n}{a_n}\right|\le \frac{\frac{1}{\beta}+\varepsilon}{\frac{1}{\alpha}-\varepsilon}\,\cdot \,\left|\frac{b_{n-1}}{a_{n-1}}\right|\le \cdots \le \left(\frac{\frac{1}{\beta}+\varepsilon}{\frac{1}{\alpha}-\varepsilon}\right)^{n-n_0}\,\cdot \,\left|\frac{b_{n_0}}{a_{n_0}}\right| $$

In particular, if $\alpha<\beta$, set $\varepsilon:=\frac{1/\alpha-1/\beta}{3}$ then $$ \frac{\frac{1}{\beta}+\varepsilon}{\frac{1}{\alpha}-\varepsilon}<1 \implies \lim_{n\to \infty}\left|\frac{b_n}{a_n}\right|=0. $$

Only if. If $\alpha \ge \beta$ then set $a_n:=\alpha^{-n}$ and $b_n:=\beta^{-n}$ for all $n$. Hence $$ \left|\frac{b_n}{a_n}\right|= \left(\frac{\alpha}{\beta}\right)^n\ge 1 $$ for all $n$, so it does not converge to $0$.

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  • $\begingroup$ sorry that i correct the problem to $a_n/b_n$ $\endgroup$ – sinoky Sep 2 '17 at 15:12
  • $\begingroup$ Just replace $\alpha \mapsto \beta$ and $\beta \mapsto \alpha$ : the same reasoning applies. $\endgroup$ – Paolo Leonetti Sep 2 '17 at 15:14

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