Find $\lim\limits_{n\rightarrow \infty} |\frac{a_n}{b_n}|$ 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.

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 $0$.
