I am currently studying mathematical analysis, specifically the theory of sequences, and I have come across a challenging problem in my homework that I am unable to solve. I would greatly appreciate any assistance or insight you can provide.
The problem is as follows:
Consider the sequence defined recursively by $x_{n+2}=x_{n+1}+\frac{x_n}{\ln x_n}$ for $n\geq 0$, where $x_0,x_1>1$. I am tasked with finding the limits of two expressions. Firstly, I need to determine $$c:=\lim_{n\rightarrow +\infty}\frac{\ln^2 x_n}{n}.$$ Secondly, I need to calculate $$\lim_{n\rightarrow +\infty}\frac{\ln^2x_n-cn}{\sqrt{n}}.$$
You have by an easy induction that $x_n>1$ for any $n\in\mathbb N$, then $(x_n)$ is increasing and it's going to infinity otherwise its limit $l$ would be such that $l=l+\frac{l}{\ln(l)}$ hence $l=0$ : contradiction.
Now you can use that to get $x_{n+2}\sim x_{n+1}$ and from $\displaystyle \frac{x_{n+2}}{x_{n+1}}=1+\frac{x_n}{x_{n+1}\ln(x_n)}$ you also have $\ln(x_{n+2})-\ln(x_{n+1})\sim \frac{x_n}{x_{n+1}\ln(x_n)}\sim\frac{1}{\ln(x_n)}$ and $\ln(x_{n+2})+\ln(x_{n+1})\sim 2\ln(x_{n+1})$ hence $\ln^2(x_{n+2})-\ln^2(x_{n+1})\sim 2$. Then $\ln^2(x_n)\sim 2n$ and hence $c=2$. For the other limit try to use first order expansions instead of equivalences as I did.