Suppose $a_1=2,a_2=3$ and $a_{n+2}=a_{n+1}+\dfrac{1}{\ln a_n}$ for every $n\ge 1$. Find $\lim\limits_{n \to \infty}\dfrac{a_n \ln a_n}{n}.$
First, we can prove $a_n>1$, which is easy by induction. Therefore, $a_{n+2}=a_{n+1}+\dfrac{1}{\ln a_n}>a_{n +1}$, which implies $\{a_n\}$ is increasing. Hence, $\{a_n\}$ either converges to a finite limit, or diverges to the positive infinity. But the former one is impossible. Ortherwise, we assume $\lim\limits_{n \to \infty} a_n=a<+\infty$, then $a=a+\dfrac{1}{\ln a}$ by taking the limits of the recursive equility. That equation has no real solution. Therefore, we can claim $a_n \to +\infty(n \to \infty).$
How to go on with this?
The hints from @Mindlack and @PhoemueX are very useful, and I'll try to give you a full solution here:
Note that $a_n\to +\infty\ (n\to\infty)$, we have $$ \begin{aligned} \ln a_{n+2} &= \ln\left(a_{n+1}+\dfrac{1}{\ln a_n}\right) \\ &= \ln a_{n+1}+\ln\big(1+\dfrac{1}{a_{n+1}\ln a_n}\big) \\ &= \ln a_{n+1}+\dfrac{1}{a_{n+1}\ln a_n}+o\Big(\dfrac{1}{a_{n+1}\ln a_n}\Big) \end{aligned} $$ so $\dfrac{\ln a_{n+2}}{\ln a_{n+1}} = 1+o(1)$, and then $\dfrac{\ln a_{n+2}}{\ln a_n} = 1+o(1)$ follows. Thus, $$ \begin{aligned} a_{n+2}\ln a_{n+2} &= a_{n+1}\ln a_{n+2} + \dfrac{\ln a_{n+2}}{\ln a_n} \\ &= a_{n+1}\ln a_{n+1} + \dfrac{1}{\ln a_n} + o\Big(\dfrac{1}{\ln a_n}\Big) + 1 + o(1) \\ &= a_{n+1}\ln a_{n+1} + 1 + o(1), \end{aligned} $$ according to Stolz Theorem, it follows that $$ \lim_{n\to\infty} \dfrac{a_n\ln a_n}{n} = \lim_{n\to\infty} \big( a_{n+2}\ln a_{n+2}-a_{n+1}\ln a_{n+1} \big) = 1. $$