Let $\{y_n\}$ be a sequence of positive reals satisfying the following recursive formula:
$$y_{n+1} = \frac{1}{3} \left(y_n + \frac{2}{y_n}\right).$$
Now, I am interested in finding out a closed form for the above recurrence, but am unable to do so. Here is what I have tried so far -
First, I tried using the method of fixed points. I solved the equation $\lambda=\frac{1}{3} \left(\lambda+\frac{2}{\lambda}\right),$ obtaining $\lambda= \pm 1$. Then, we can deduce the following:
$$y_{n+1} -1 = \frac{1}{3}\left(y_n-1\right) + \frac{2}{3} \left(\frac{1}{y_n}-1\right) \iff \frac{y_{n+1}-1}{y_n-1}=\frac{y_{n}-2}{3y_n},$$
and: $$y_{n+1} + 1 = \frac{1}{3}\left(y_n+1\right) + \frac{2}{3} \left(\frac{1}{y_n}+1\right) \iff \frac{y_{n+1}+1}{y_n+1}=\frac{y_{n}+2}{3y_n}$$
Thus, we have the following:
$$\frac{y_{n+1}+1}{y_{n+1}-1}= \frac{y_{n}+2}{y_{n}-2} \cdot \frac{y_{n}+1}{y_{n}-1}$$
After this, I am stuck, especially since I can't seem to do anything with the expression $\frac{y_{n+2}}{y_{n-2}}$.
I also tried to use Generating functions but to no avail. Setting $f(x)=\sum_{n=0}^{\infty} y_n x^n$, I only managed to obtain the following:
$$\sum_{n=0}^{\infty} y_n x^n = y_0 + \sum_{n=1}^{\infty} \left(\frac{1}{3} y_{n-1} + \frac{2}{3y_{n-1}}\right)x^n$$
$$=y_0 + \frac{1}{3} x \sum_{n=0}^{\infty} y_n x^n + \frac{2}{3} \sum_{n=1}^{\infty} \frac{1}{y_{n-1}}x^{n} $$
$$=a_0+\frac{1}{3} x f(x)+ \frac{2}{3} \sum_{n=1}^{\infty} \frac{1}{y_{n-1}}x^{n}$$
At which point, I have absolutely no clue on how to proceed.
Any help would be much appreciated!