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?

  • $\begingroup$ u_(n+1) is very close to the harmonic mean of 2^n and u_n when u_n/2^n is small. Perhaps that helps for another derivation? Gerhard "Ask Me About System Design" Paseman, 2012.10.26 $\endgroup$ – Gerhard Paseman Oct 26 '12 at 22:51

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

  • $\begingroup$ Your first display is valid even when $u_n=0$, and this case occurs if and only if $\lambda=0$. $\endgroup$ – GH from MO Oct 26 '12 at 22:20
  • $\begingroup$ that's very nice $\endgroup$ – Pietro Majer Oct 26 '12 at 22:37
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    $\begingroup$ So $x=u_n$ is a solution of $(1+\frac{x}{N})^N=1+\lambda$ (with $N:=2^n$), which we may see as an approximated problem for $e^x=1+\lambda$... $\endgroup$ – Pietro Majer Oct 27 '12 at 7:47
  • $\begingroup$ @Pietro: that´s a nice way to look at it. $\endgroup$ – Ramiro de la Vega Oct 27 '12 at 18:46

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