Let $(a_n)_{n\ge1}$ a sequence of real positive numbers such that $a_n+\frac{n}{a_{n+1}^2} \le a_{n+1} \le \frac{a_{n-1}^2}{a_n}+\frac{n+1}{a_n^2}$ for any $n \ge 1$. Prove that $b_n = \frac{a_n}{\sqrt[3]{n^2}}$ is a convergent sequence and find its limit.
First of all, it's clear that $a_n$ is a increasing sequence from the first inequality given. So $a_n$ has a limit and it is either a finite number or infinity. If we suppose that $a_n$ is convergent and has a finite limit (let it be $l$), by using limit in the first inequality we obtain: $l+\infty \le l$, which is false and it means that $\lim_{n \to \infty} a_n = \infty$, so we may apply Cesaro-Stolz for finding the limit of $b_n$, but here I didn't know how to continue. Can you help me?
Best Answer
We have $a_n$ increasing and positive, and $$ a_{n} \leq a_{n+1} - \frac{n}{a_{n+1}^2}, $$ Thus $a_{n+1}^3 \geq n$, and $$ \frac{a_{n-1}^2}{a_n} + \frac{n+1}{a_n^2} \leq a_n + \frac{(n-1)^2}{a_n^5} + \frac{3-n}{a_n^2}, $$ and using $a_{n}^3 \geq (n-1)$ $$ \frac{(n-1)^2}{a_n^5} + \frac{3-n}{a_n^2} \leq \frac{(n-1)^2}{a_n^5} \left(1-\frac{n-3}{(n-1)^2}a_n^3 \right) \leq 2 \frac{(n-1)}{a_n^5} < 2\frac{1}{(n-1)^{2/3}} $$ so altogether $$ a_n + \frac{n}{a_{n+1}^2}\leq a_{n+1} \leq a_n + \frac{2}{(n-1)^{2/3}} . $$ So $$ \sum_{k=2}^n a_{k+1} -a_{k} \leq 2\sum_{k=2}^n k^{-2/3} \leq 2 \int_{1}^{n-1} \frac{1}{t^{2/3}} dt = 6 n^{1/3}-6, $$ So we have obtained that $$ (n-1)^{1/3}\leq a_n \leq 6(n-1)^{1/3}+C, $$ and the conclusion is clear.