In my research I came upon a recursively defined sequence, and I'm pretty sure it converges to $\sqrt{2}$ though I can't prove it easily. I don't think it is a difficult question but I'm not sure.


Consider the following sequence of functions over $\mathbb{R}$, where it makes sense:


$f\_0(x)=0$,


$\displaystyle f\_{n+1}(x)=\frac{1}{2(x-f\_n(x))}\hspace{1cm}$ ($n\geq 0)$.


Now let us define the sequence $(x\_n)$ by $\displaystyle x\_n:=\max\left\{y\in[0,\sqrt{2}[\,:\quad y=\frac{1}{2y}+f\_n(y)\right\}$.



> 
> Question: is is true that $x\_n\rightarrow\sqrt{2}$ when $n\rightarrow +\infty$?
> 
> 
> 


Numerical evidence strongly suggest that, and it completely makes sense with the problem it originated from. The issue is that the functions $f\_n$ have more and more poles as $n$ grows, and there is no function it converges to. It looks like the set of the poles of $f\_n$ tends to be dense in $[-\sqrt{2},\sqrt{2}]$ when $n\rightarrow +\infty$, and $f\_n$ is always decreasing outside of the poles. The poles seem to accumulate more around $\pm\sqrt{2}$ than around $0$.


For visual reference, one can see a graph of $f\_{10}$ [here](https://www.wolframalpha.com/input/?i=1%2F(2(x-1%2F(2(x-1%2F(2(x-1%2F(2(x-1%2F(2(x-1%2F(2(x-1%2F(2(x-1%2F(2(x-1%2F(2(x-1%2F(2x))))))))))))))))))))


In advance, thank you for your interest/time.


Edit: I added the fact that I'm only interested in the $y\in[0,\sqrt{2}[$. I don't care what happens outside the interval since then it's trivial.



Maximum zero converges to $\sqrt{2}$
Modify the function so the equation becomes $f\_n(x)=0$. That is


$$f\_0(x)=\frac{1}{2 x}-x \quad\quad\quad f\_n(x)=\frac{1}{2 \left(\frac{1}{2 x}-f\_{n-1}(x)\right)}-x+\frac{1}{2 x}$$


Multiply both side of the equation by $-U\_n(\sqrt{1/2}x)x$


$$f\_0(x)=-U\_0(\sqrt{1/2}x)x\left(\frac{1}{2 x}-x\right) \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{ } \\ f\_n(x)=-U\_n(\sqrt{1/2}x)x\left(\frac{1}{2 \left(\frac{1}{2 x}-\frac{f\_{n-1}(x)}{-U\_{n-1}(\sqrt{1/2}x)x}\right)}-x+\frac{1}{2 x}\right)$$


Substitute $x\to\sqrt{2}x$ and make common denominator


$$f\_0(x)=U\_2(x) \quad\quad\quad f\_n(x)=\frac{U\_n(x) \left(\left(1-4 x^2\right) f\_{n-1}(x)+U\_{n-1}(x)\right)}{f\_{n-1}(x)+U\_{n-1}(x)}$$


Replace $f\_{n-1}$ by $U\_{n+1}(x)$ which is valid for $n=1$ and hopefully more.
Also replace $U\_{n-1}(x)$ using the recurrence relation for $U\_{n+1}(x)$, and expand


$$f\_0(x)=U\_2(x) \quad\quad\quad f\_n(x)=\frac{U\_n(x) \left(\left(1-4 x^2\right) U\_{n+1}(x)+2 x U\_n(x)-U\_{n+1}(x)\right)}{U\_{n+1}(x)+2 x U\_n(x)-U\_{n+1}(x)}
\\f\_0(x)=U\_2(x) \quad\quad\quad f\_n(x)=U\_n(x)-2 x U\_{n+1}(x)= U\_{n+2}(x)\hspace{3.8cm}\text{ } $$


Now it's obvious that the maximal root approaches 1 which correspond to $\sqrt{2}$ because of the substitution.


The multiplication just cancel out singularities, but even if it did introduce additional roots, these would be in the range $(0,1)$ which doesn't affect the limit of the maximal root.


