Finding a sequence which $\to 0$ "fast enough" In what follows let the sequences $\{a_n\}$ and $\{b_n\}$ be real numbers strictly between 0 and 1 satisfying:
$$ a_n \downarrow 0 \textrm{ and } b_n \downarrow 0 $$
I am trying to figure out whether it is possible to find such a sequence $\{a_n\}_{n \geq 0}$ with the property that, for any other such sequence $\{b_n\}_{n \geq 0}$, there always exists $N$ (possibly depending on $\{b_n\}$) so that:
$$ a_N > C \cdot \frac{(b_N)^{\alpha}}{b_{N+1}}, \quad \alpha > 1 $$
where $C$ is some positive constant depending only on $\alpha$ (I think this is enough to make the question make sense, but I might be wrong).
I've checked this for some easy progressions like $a_n = n^{-\beta}$ for some $\beta >1$, and it's not hard to come up with some sequences $b_n$ that don't work. I think for this specific example the sequence $b_n = n^{-\gamma}$ with something like $\gamma > \beta/(\alpha-1)$ provides a counterexample.

Suppose towards a contradiction that it is true.  Then there is $(a_n)_{n=0}^\infty$ such that for each $\alpha>1$ (or, for that matter, for some $\alpha>0$), we can find $C_\alpha>0$ with the desired property.  Now, select any $b_0\in(0,1)$, and for $n\in\mathbb{N}$, put
\begin{equation}\displaystyle b_{n+1}=\frac{C_\alpha b_n^\alpha}{a_n}\wedge\frac{1}{2}b_n.\end{equation}
Then $(b_n)_{n=0}^\infty\downarrow 0$ in $(0,1)$.  On the other hand, for every $n\in\mathbb{N}$ we have
\begin{equation}b_{n+1}\leq\frac{C_\alpha b_n^\alpha}{a_n}\end{equation}
and hence
\begin{equation}a_n\leq\frac{C_\alpha b_n^\alpha}{b_{n+1}}.\end{equation}
