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: For all $n\ge 1$ is
$$
 x_{n+1} = \frac{2n}{2n-1} \cdot \frac 12 \left( x_n + \frac{2}{x_n} \right) \ge \frac{2n}{2n-1}\sqrt 2 \,,
$$ using the inequality between the arithmetic and the geometric mean. So we have
$$
 x_n^2 \ge 2 \left( \frac{2n-2}{2n-3}\right)^2
$$
for $n \ge 2$. It follows that for $n\ge 2$,
$$
x_{n+1} = \frac{n}{2n-1}\frac{x_n^2+2}{x_n}
 \le \frac{n}{2n-1} \frac{x_n^2 + \left( \frac{2n-3}{2n-2}\right)^2x_n^2}{x_n} = a_n x_n 
$$
with
$$
a_n := \frac{n}{2n-1} \left( 1+ \left( \frac{2n-3}{2n-2}\right)^2\right) 
= \frac{8 n^3 - 20n^2+13n}{8n^2-20n^2+16n - 4} < 1 \, .
$$
This shows that $(x_n)_{n \ge 2}$ is decreasing. It is also bounded below (by zero) and therefore convergent. Now you can take the limit in the recurrence relation and show that the limit is $\sqrt 2$.
A: 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.
A: We will prove that
$\sqrt2<x_{n+1}<x_n\;\;$ for any $\;n\in\Bbb N\;\land\;n\geqslant2\;.\quad\color{blue}{(*)}$
For any $\;n\in\Bbb N\;$ it results that
$\begin{align}x_{n+1}&=\dfrac n{2n-1}\!\cdot\!\dfrac{x_n^2+2}{x_n}=\dfrac n{2n-1}\!\cdot\!\dfrac{\left(x_n-\sqrt2\right)^2+2\sqrt2x_n}{x_n}\geqslant\\&\geqslant\dfrac{2\sqrt2\,n}{2n-1}>\sqrt2\;.\end{align}$
In particular we get the first inequality of $\,(*)\,$ and we also obtain that
$x_n\geqslant\dfrac{2\sqrt2\,(n-1)}{2n-3}\;\;$ for any $\;n\in\Bbb N\;\land\;n\geqslant2\;.\quad\color{blue}{(1)}$
Moreover, for any $\;n\in\Bbb N\;\land\;n\geqslant2\;,\;$ it results that
$2n-(n-1)x_n^2\underset{\color{brown}{(1)}}{\leqslant\;}2n-\dfrac{8(n-1)^3}{(2n-3)^2}=\dfrac{-6n+8}{(2n-3)^2}<0\,.\;\;\color{blue}{(2)}$
For any $\;n\in\Bbb N\;\land\;n\geqslant2\;,\;$ it also results that
$\begin{align}x_{n+1}&=\dfrac n{2n-1}\!\cdot\!\dfrac{x_n^2+2}{x_n}=\dfrac{\big[(2n-1)-(n-1)\big]x_n^2+2n}{(2n-1)x_n}=\\&=\dfrac{2n-(n-1)x_n^2+(2n-1)x_n^2}{(2n-1)x_n}\underset{\color{brown}{(2)}}{<}x_n\;.\end{align}$
So we have proved the inequalities $\,(*)\,$, consequently the sequence $(x_n)$ is bounded and eventually decreasing, therefore there exists $\lim\limits_{n\to\infty}x_n=l\in\Bbb R^+\,.$
Moreover ,
$l=\lim\limits_{n\to\infty}x_{n+1}=\lim\limits_{n\to\infty}\left(\dfrac n{2n-1}\!\cdot\!\dfrac{x_n^2+2}{x_n}\right)=\dfrac{l^2+2}{2l}\;,$
hence ,
$l=\dfrac{l^2+2}{2l}\quad,$
$2l^2=l^2+2\quad,$
$l^2=2\quad,$
$l=\sqrt2\,.$
It means that $\;\lim\limits_{n\to\infty}x_n=\sqrt2\,.$
A: In general, you are given $x_{n+1} = f_n(x_n)$.  You wanted to show $f_n(x)-x \leq 0$ for all $x \geq \sqrt{2}$, but that's not true.
However, what you have shown is that $(x_n)$ can only move away from $\sqrt{2}$ when it's very close to $\sqrt{2}$.  If we ignore for a moment whether it's even possible for $x_n$ to get that close, the question this would raise for me is "how large can the move away from $\sqrt{2}$ even be?"  More precisely, what is the maximum of $f_n(x) - x$ on $\sqrt{2} \leq x \leq \sqrt{2n/(n-1)}$.
Answering this is a simple question of optimization (at least by derivatives if not algebra).  After a bit, we find that maximum is $\sqrt{2}/(2n-1)$ (at $x=\sqrt{2}$), which goes to zero in the limit.
And that would finish it.  You know that outside a diminishing neighborhood of $\sqrt{2}$, the sequence is always decreasing, and if it even were possible to increase, the maximum increase diminishes to $0$.
