The following is not exactly a research question (it was originated from manufacturing of exercises for calculus), and has no other motivation than explaining a phenomenon. I apologize if it is inappropriate (and will quickly remove it). 


Consider the sequence of real numbers defined recursively as follows
$$u\_0:=\lambda > -1\\ ,$$ 
$$u\_{n+1}=\frac{2u\_n}{1+\sqrt{1+2^{-n}u\_n}}\\ .$$ 
Numerical evidence suggests that $u\_n$ always converges to $\log(1+\lambda)$. I imagine there should be a simple explanation. How can one prove it?



Computing the limit of a certain recursively defined sequence 
Multiplying top and bottom by the conjugate and simplifying (assuming $u\_n \neq 0$) we get: $$u\_{n+1}=2^{n+1}(\sqrt{1+2^{-n}u\_n}-1).$$ Calling $v\_n:=\frac{u\_n}{2^n}+1$ we have the recursion: $v\_{n+1}=\sqrt{v\_n}$ and therefore $v\_n=v\_0^{2^{-n}}$. Going back to $u$´s we have $u\_n=2^n[(1+\lambda)^{2^{-n}}-1]$. Finally: $$\lim\_{n\to\infty} u\_n=\lim\_{h\to 0}\frac{(1+\lambda)^h-(1+\lambda)^0}{h}= \ln (1+\lambda).$$


