# A sequence of nested fractions with a counter-intuitive limit

Given $a,b\in\mathbb C$, let us construct the following sequence:

\begin{align} a+b&=a+b\\ \cfrac a{a+b}+\cfrac b{a+b}&=1\\ \cfrac a{\cfrac a{a+b}+\cfrac b{a+b}}+\cfrac b{\cfrac a{a+b}+\cfrac b{a+b}}&=a+b\\ \cfrac a{\cfrac a{\cfrac a{a+b}+\cfrac b{a+b}}+\cfrac b{{\cfrac a{a+b}+\cfrac b{a+b}}}}+\cfrac b{\cfrac a{{\cfrac a{a+b}+\cfrac b{a+b}}}+\cfrac b{{\cfrac a{a+b}+\cfrac b{a+b}}}}&=1\\ &\vdots\\ L(a,b)&=\;? \end{align}

Since the sequence keeps oscillating between only two numbers, the limit $L(\cdot\,,\cdot)$ doesn't seem to exist, but I saw a claim that $L(1,3)=2$. This could be a hint that it converges to $(a+b)/2$ or $\sqrt{a+b}$ …

So, what is going on here?

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What is L() defined as? –  Victor Mar 17 '12 at 1:10
It does not converge, as you observed. The claim was incorrect. –  Alex Becker Mar 17 '12 at 1:11
It seems that $L(a,b)$ only exists when $a + b = 1$, so the claim is likely incorrect. –  JavaMan Mar 17 '12 at 1:14
Continuous fractions is a rather specific form of limiting procedure, rather unconnected to the one you are asking about. –  Mariano Suárez-Alvarez Mar 18 '12 at 20:35

The two terms on the left of each equation have the same denominator, so the numerators can be added. You can define $f_1(a,b)=a+b$ and $f_n(a,b)=\frac {a+b}{f_{n-1}(a,b)}$. Then it should be clear that $$f_n(a,b)=\begin {cases} a+b & n \text{ odd} \\1 & n \text { even} \end {cases}$$ This doesn't have a limit unless $a+b=1$. It is similar to asking what $\lim_{n \to \infty} \sum_{i=1}^n (-1)^i$ is. It oscillates between two values, so there is no limit.

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The claim is from a YouTube video. Someone replied to me there. Here's what they said, with some persnickety details I added.

First of all, this question doesn't even make sense if $a+b$ is zero (the second line would be the absurd sentence $a/0-a/0=1$), so let's impose that it isn't.

$L(a,b)$ should be interpreted as a fractal fraction, so that we can write

$$L(a,b) = \frac{a}{L(a,b)} + \frac{b}{L(a,b)} = \frac{a+b}{L(a,b)}\quad,$$

thus

$$L(a,b)^2 = a+b\quad.$$

This suggests there are two distinct solutions, but the limit has to be unique.

$$L(a,b) = \sqrt{a+b}$$

works if $a+b\in\mathbb R^*_+$, so it must be answer. I wonder what stops $-\sqrt{a+b}$ from being the value of the limit in the general case $a+b\in\mathbb C^*$.

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«Fractal fraction» does not seem to mean anything... –  Mariano Suárez-Alvarez Mar 18 '12 at 20:33
All you've proven is that the limit, if it exists, must equal $\sqrt{a+b}$ or $-\sqrt{a+b}$. However, the limit might not exist at all... –  Ilmari Karonen Mar 18 '12 at 20:49
Maybe you could give a link to the video? –  Gerry Myerson Mar 19 '12 at 5:58
@Gerry Good idea. youtube.com/watch?v=GFLkou8NvJo#t=1m And there is the spoiler that wau is 1. –  Luke Mar 20 '12 at 1:56