Question

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?

Answer 1

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: $$\lim_{n\to\infty} u_n=\lim_{h\to 0}\frac{(1+\lambda)^h-(1+\lambda)^0}{h}= \ln (1+\lambda).$$