I'm trying prove that $\displaystyle \limsup_{n\to\infty}\frac{a_n}{b_n}$ is finite, when I know that $\displaystyle \frac{a_{n+1}}{a_n}≤\frac{b_{n+2}}{b_n}$.
I'm a little stuck, but have some inclination to think that it has something to do with manipulating $\displaystyle \frac{a_n}{b_n}≤\frac{a_{n-2}}{b_{n-2}}\frac{b_{n+1}}{b_{n-1}}$, which follows from the above inequality.
Thanks for any help!
Edit: Assume $a_n$ and $b_n$ are positive sequences.
This is false as stated. Take $a_n=n^3$ and $b_n=n^2$. Then $\limsup \frac{a_n}{b_n}=+\infty$ and yet $$ \frac{b_{n+2}}{b_n}-\frac{a_{n+1}}{a_n}=\left(1+\frac{2}{n}\right)^2-\left(1+\frac{1}{n}\right)^3=\frac{1}{n}\left(1+\frac{1}{n}-\frac{1}{n^2}\right)\geq\frac{1}{n}\geq 0\qquad\forall n\geq 1. $$
If you replace your condition by $$ \frac{a_{n+1}}{a_n}\leq \frac{b_{n+1}}{b_n}\quad\Rightarrow\quad \frac{a_{n+1}}{b_{n+1}}\leq \frac{a_n}{b_n} $$ then the sequence is nonincreasing. So $$-\infty\leq \lim \frac{a_n}{b_n} = \limsup \frac{a_n}{b_n}\leq \frac{a_1}{b_1}<\infty. $$