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How can I solve for $x$:

$$1=\cfrac{1}{x}+\cfrac{1}{1+\cfrac{1}{x}}+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{x}}}+\cdots$$

Any clues?

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does the rhs converge for any $x$? –  yoyo Jun 14 '13 at 18:16
    
Some quick calculations lead me to believe that the series does not converge, but I might be wrong. –  Javier Badia Jun 14 '13 at 18:24
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Each term can be written as $\frac{a+bx}{c+dx}$ where $a,b,c,d$ are Fibonacci numbers. I'll need a little moment to figure out which they are. –  Thomas Andrews Jun 14 '13 at 18:26
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@ThomasAndrews $\displaystyle \frac{F_nx+F_{n-1}}{F_{n+1}x+F_n}$ starting at $n=0$ I believe. (typo) –  anon Jun 14 '13 at 18:27
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2 Answers

$$a_{n+1} = \frac{1}{1+a_n},\ a_0=1/x,\ \sum_{n=0}^{\infty}a_n=1$$

Unfortunately, $a_n$ converges to a non zero constant $\frac{\sqrt{5}-1}{2}$, so that the sum of all $a_n$'s is infinity.

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We deal with a function $x\mapsto \frac{1}{1+\frac{1}{x}}$, which if we take $\frac{1}{x}$ as argument, we should rather write as $x\mapsto \frac{1}{1+x}$.

Iterated, this gives the fixed point iteration for finding a soltion of $\frac{1}{1+x}=x$. And therefore all your terms eventually turn out converge to a solution of $1=(1+x)\ x$, namely $\frac{\sqrt{5}-1}{2}$ (see golden ratio), making the total sum diverge.

Some Mathematica code:

enter image description here

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+1 Beat me to it :) –  nbubis Jun 14 '13 at 18:33
    
Do the terms ever converge to $-\varphi$ besides $x=-\varphi$? –  anon Jun 14 '13 at 18:34
    
@anon: $|\varphi|\approx 0.618<1$, so the result is positive. There is no square, the $x$ on the right hand side of $\frac{1}{1+x}=x$ is dropped. –  NiftyKitty95 Jun 14 '13 at 18:42
    
@vermiculus: eww, why did you do that? ;P –  NiftyKitty95 Jun 15 '13 at 0:03
    
@NickKidman Hehehehehe Hey man, if people vote it up, it ought to be persistent. It'd be nice to see a formal proof, even though the original equation is a definition for $\phi$, but a table --- while not conclusive --- has immediate and observable value. (What would be better to see is a graph of these data.) –  vermiculus Jun 15 '13 at 0:06
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