Prove that the sequence $\{u_n\}$ is defined by $0\lt u_1\lt u_2 \;\text {and} \;u_{n+2}=\sqrt {u_{n+1}u_n}\; \text{for}\; n\ge 1$, converges to $\sqrt[3]{u_1{u_2}^2}$.
I did this using subsequences, I posted below. This kind of problems are usually done by monotonicity and boundedness. So, I suspect if their is a solution to this problem using this method.
I answered this in this method here.
Hint : we have $u_n>0$ for every $n$, by induction. Then $v_n={\ln}(u_n)$ is a linear recurrent sequence : it satisfies $v_{n+2}=\frac{v_{n+1}+v_n}{2}$. The characteristic polynomial is $X^2-\frac{X+1}{2}$, whose roots are $1$ and $-\frac{1}{2}$. So there are two constant $a$ and $b$ such that $v_n=a\times 1+b\times \bigg(\frac{-1}{2}\bigg)^n$. Hence $u_n=e^{a+b\times \big(\frac{-1}{2}\big)^n}$.
Then $(u_n)$ converges to $e^a$, and it is easy to compute that $\sqrt[3]{u_1u_2^2}=e^a$.
