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?
 A: 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.
A: 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 $a+b\in\mathbb C^*$.
