Question:
let $\alpha\in (0,1)$,and the postive sequence $\{a_{n}\}$ such $$\lim_{n\to\infty}\inf \left(n^{\alpha}\left(\dfrac{a_{n}}{a_{n+1}}-1\right)\right) =\lambda\in (0,+\infty)$$
show that $$\lim_{n\to\infty}n^k a_{n}=0,k>0$$
maybe this is Ratio Test and Its Generalizations?
This problem is The Chinese Mathematics Competitions 2014 before,But I can't prove it.can you help
Let $\mu\lt\lambda$, $\mu$ positive, then, for every $n$ large enough, say, for every $n\geqslant k$, $$n^{\alpha}\left(\frac{a_{n}}{a_{n+1}}-1\right)\geqslant\mu,$$ hence $$a_{n+1}\leqslant a_n\,\left(1+\frac{\mu}{n^\alpha}\right)^{-1}.$$ This implies that $a_n\leqslant C/b_n$, where $$b_n=\prod_{i=1}^n\left(1+\frac{\mu}{i^\alpha}\right).$$ For every $x$ small enough, $1+x\geqslant\mathrm e^{x/2}$ hence, for every $n$ large enough, $$b_n\geqslant C'\exp\left(\frac\mu2\sum_{i=1}^n\frac1{i^\alpha}\right)\geqslant C'\exp\left(\frac\mu2\int_1^{n+1}\frac{\mathrm dt}{t^\alpha}\right)\geqslant C''\exp\left(\frac\mu2\frac{n^{1-\alpha}}{1-\alpha}\right). $$ Finally, there exists $c$ and $\nu$ positive such that, for every $n$, $$a_n\leqslant c\exp(-\nu n^{1-\alpha}),$$ and this suffices to conclude.
