Find limit of sequence $(x_n)$ Find limit of sequence $(x_n)$:
$$x_1 = a >0$$
$$x_{n+1} = \frac{n}{2n-1}\frac{x_n^2+2}{x_n}, n \in Z^+$$
I think I can prove $(x_n)$ is low bounded (which is obvious that $x_n>0$) and decreasing sequence. Then I can calculate the limit of sequence is $\sqrt{2}$
All my attempts to prove it's a decreasing sequence have been unsuccessful.
My attemps: Try to prove $x_{n+1}-x_{n} <0$ from a number $N_0$ large enough. It lead to I have to prove $x_n \ge \sqrt{\frac{2n}{n-1}}$ and I stuck.
Does anyone have an idea?

A recursion shows that the $x_n$ for $n \ge 2$ are well-defined and bounded below by $\sqrt{2}$, since $x^2+2-2\sqrt{x} = (x-\sqrt{2})^2$ is always positive. Set $y_n = x_n-\sqrt{2}$. Then for $n \ge 2$,
\begin{eqnarray*}
y_{n+1} 
&=& \frac{n(x_n^2+2)-\sqrt{2}(2n-1)x_n}{(2n-1)x_n} \\
&=& \frac{n(x_n-\sqrt{2})^2+\sqrt{2}x_n}{(2n-1)x_n} \\
&=& \frac{ny_n^2}{(2n-1)x_n} + \frac{\sqrt{2}}{2n-1} \\
&\le& \frac{n}{2n-1}y_n+ \frac{\sqrt{2}}{2n-1},
\end{eqnarray*} since $y_n \le x_n$.
For $n \ge 2$, $n/(2n-1) \le 2/3$, so
$$y_{n+1} \le \frac{2}{3}y_n+\frac{1}{3}z_n, \text{ where } z_n = \frac{3\sqrt{2}}{2n-1}$$
Given $\epsilon>0$, one can find an integer $N$ such that for every $n \ge N$, $z_n \le\epsilon$ so $$y_{n+1}-\epsilon \le \frac{2}{3}(y_n-\epsilon),$$
so by recursion
$$y_n-\epsilon \le \Big(\frac{2}{3}\Big)^{n-N}(y_N-\epsilon).$$
As a result, $\limsup_{n \to +\infty} y_n \le \epsilon$. Since this is true for every $\epsilon>0$, we get the desired conclusion.
