A problem about the limit of a sequence

Question

Suppose that $\alpha\in(0,1)$ and let $\{a_n\}_{n\geq 1}$ be a sequence with positive terms satisfying $$\liminf_{n\to\infty} \;\left(n^{\alpha}\left(\frac{a_n}{a_{n+1}}-1\right)\right)=\lambda\in(0,+\infty).$$ Show that $\lim\limits_{n\to\infty}n^ka_n=0$ for all $k>0$.

Answer 1

Hint. By definition of $\liminf$, for $\epsilon=\lambda/2>0$ there is $N\geq 1$ such that for $n\geq N$ $$n^{\alpha}\left(\frac{a_{n}}{a_{n+1}}-1\right)\geq \lambda-\epsilon=\lambda/2\implies \frac{a_{n+1}}{a_n}\leq \frac{1}{1+\frac{\lambda}{2n^{\alpha}}}.$$ Now for $n\geq N$, $$0<a_{n+1}=\frac{a_{n+1}}{a_n}\cdots\frac{a_{N+1}}{a_N}\cdot a_N\leq \frac{a_N}{\left(1+\frac{\lambda}{2n^{\alpha}}\right)^{n+1-N}}.$$ Note that, as $n\to+\infty$, $$\left(1+\frac{\lambda}{2n^{\alpha}}\right)^{n+1-N}\sim \exp\left(n^{1-\alpha}\lambda/2\right).$$