Limit of recursive sequences: $x_{n+1}=\frac 1 3 (x_n+2y_n), y_{n+1}=\frac 1 3 \left( \frac 1 {x_n}+ \frac 1 {2y_n} \right).$ Let $\left(\,x_{n}\,\right)_{\ n\ \geq\ 1}$ and
$\left(\,y_{n}\,\right)_{\ n\ \geq\ 1}$ be two sequences defined as follows
$\left(\,x_{1},y_{1} > 0\,\right)$:
$$
\left\{\begin{array}{rcl}
{\displaystyle x_{n + 1}} & {\displaystyle =} &
{\displaystyle{1 \over 3}\left(\,x_{n} + 2y_{n}\,\right)}
\\[2mm]
{\displaystyle y_{n + 1}} & {\displaystyle =} &
{\displaystyle{1 \over 3}\left(\,{1  \over x_{n}} + {1  \over 2y_n}\,\right)}
\end{array}\right.
$$
Show that the sequences are convergent and find their limits.
MY TRY:


*

*I think about using the Weierstrass' Theorem ( every bounded and monotone sequence is convergent ), but I can't find the monotonity for any sequence. 

*Afterwards, I can think of denoting the two limits as $L_{1}$ and $L_{2}$ and replacing them in the reccurence relationship.

 A: Let us consider the two sequences
$$a_n = \frac{x_n}{y_n}$$
and
$$b_n = x_n y_n$$ instead
Dividing the two given recurrences (of $x_{n+1}$ and $y_{n+1}$) gives us
$$a_{n+1} = 2 b_n$$

Note that it is enough to show that $b_n$ is convergent, as that
  implies $a_n$ is convergent, which in turn implies $x_n^2 = a_nb_n$ is
  convergent, which implies $y_n$ is too.

Now multiplying the two give recurrences(of $x_{n+1}$ and $y_{n+1}$) in the question gives us
$$9 b_{n+1} = 2 + \frac{a_n}{2} + \frac{2}{a_n} = 2 + b_{n-1} + \frac{1}{b_{n-1}}$$
To simplify the algebra, Let $$s_n = \sqrt{b_n}$$ 
Thus
$$(3s_{n+1})^2 = \left(s_{n-1} + \frac{1}{s_n}\right)^2$$
Since $s_n \gt 0$, we have
$$3s_{n+1} = s_{n-1} + \frac{1}{s_{n-1}}$$
Now by considering the even $n$ and odd $n$ separately, we just need to analyze the sequence 
$$3c_{n+1} = c_n + \frac{1}{c_n}$$
(for eg by setting $c_k = s_{2k}$ or $c_k = s_{2k+1}$)
Let $$L = \frac{1}{\sqrt{2}}$$
Let $$e_n = c_n -L$$
Note that $$e_n = 0 \implies e_{n+1} = 0 \implies c_n \to L$$
So assume $e_n \ne 0$
We have that $$3L = L + \frac{1}{L}$$ and so
$$3c_{n+1} -3L = c_n - L + \frac{1}{c_n} - \frac{1}{L}$$
i.e.
$$ 3 e_{n+1} = e_n - \frac{e_n}{Lc_n}$$
Thus
$$\left|\frac{e_{n+1}}{e_n}\right| = \frac{|1 - \frac{1}{Lc_n}|}{3}$$
Let us now show that $$\frac{|1 - \frac{1}{Lc_n}|}{3} \le d$$ for some constant $d \lt 1$ which implies $e_n \to 0$.
Now we have that  $$c_n \ge \frac{2}{3}$$
using $x + \frac{1}{x} \ge 2$ (put $x = c_{n-1}$).
$$ \implies Lc_n \ge \frac{\sqrt{2}}{3}$$
And so
$$\frac{3Lc_n}{\sqrt{2}} \ge 1$$
$$\frac{(3-\sqrt{2})Lc_n}{\sqrt{2}} \ge 1 - Lc_n$$
Thus
$$\frac{3-\sqrt{2}}{3\sqrt{2}} \ge \frac{1 - Lc_n}{3Lc_n}$$
Now if $Lc_n \gt 1$, then $$\frac{\left|1 - \frac{1}{Lc_n}\right|}{3} = \frac{1 - \frac{1}{Lc_n}}{3} \le \frac{1}{3}$$
if $Lc_n \le 1$ then
$$\frac{\left|1 - \frac{1}{Lc_n}\right|}{3} = \frac{1 -Lc_n}{3Lc_n} \le \frac{3-\sqrt{2}}{3\sqrt{2}}$$
Thus $e_n \to 0$ and so $c_n \to L$. Thus $s_{2n}, s_{2n+1} \to L$ and so $b_n$ is convergent.
