Newsgroups: sci.math.research
Subject: Re: x+1 = sqrt(1+x*sqrt(1+(x+1)*sqrt(1+...)))
Date: 14 Jun 1997 01:34:34 +0100
Organization: University of Cambridge, England

Timothy Chow wrote:

> I believe Ramanujan proved that
> 
>    x+1 = sqrt(1 + x*sqrt(1 + (x+1)*sqrt(1 + (x+2)*sqrt(1 + ...))))
> 
> Is there a simple proof?

Yes. Write f(x) for your RHS.

1. We can bound f(x) below by sqrt(x*sqrt(x*sqrt(...))) = x,
   and above by sqrt((x+1)sqrt(2(x+1)sqrt(4(x+1)sqrt(...)))) = 2(x+1).

2. If f(x) lies between A.(x+1) and B.(x+1) for all x, for some
   A,B  with A<1<B (note: by the above this holds for A=1/2, B=2)
   then f(x)^2 = 1+xf(x+1) lies between 1+Ax(x+2) and 1+Bx(x+2)
   and hence between A(x+1)^2 and B(x+1)^2. Thus we can repeat
   with sqrt(A) and sqrt(B).

3. Letting the number of iterations increase without limit,
   A,B both tend to 1. Hence f(x) = x+1.

A nice proof. I wish I'd found it.

-- 
Gareth McCaughan       Dept. of Pure Mathematics & Mathematical Statistics,
gjm11@dpmms.cam.ac.uk  Cambridge University, England.