$$0<u_1<u_2 \text{ and } u_{n+2} =\sqrt{u_{n+1}. u_n}$$ Show that the limit of this sequence is ${(u_1.u_2^2)}^{1/3}$.
My attempt: By computing individual terms I see that every term can be written in terms of $u_1$and $u_2$. Also this sequence is monotonic increasing and it also has an upper bound by the AM GM inequality. But it doesnot seem to get me anywhere. Please help.
Hint : Let $\left(v_n\right)$ be the sequence defined by $v_n=u_{n}u_{n+1}^2$ then $$ v_{n+1}=u_{n+1}u_{n+2}^2=u_{n+1}\times\left(u_{n+1}u_n\right)=u_nu_{n+1}^2=v_n $$ As $v_{n+1}=v_n$ for all $n$, the sequence $\left(u_nu_{n+1}^2\right)$ is constant, so for all $n$, $v_n=v_1$ which means $$ u_nu_{n+1}^2=u_1u_2^2 $$
If you succeed in showing that the sequence $\left(u_n\right)$ converges, you have your limit : if $u_n \underset{n \rightarrow +\infty}{\rightarrow}\ell$ then you have $$ \ell\times \ell^2=u_1u_{2}^2 \Rightarrow \ell=\left(u_1u_2^2\right)^{1/3} $$
