Let $x_1 \ge 2, x_{n+1}=1+\sqrt{x_n-1}$. Find the limit, if it is convergent. Let $x_1 \ge 2, x_{n+1}=1+\sqrt{x_n-1}$. Find the limit, if it is convergent.  
My attempt: I'm trying to use the Monotone convergence theorem to show that it is convergent. So, I have to show that the sequence is decreasing and bounded below/ increasing and bounded above. 
I have been able to show that it is bounded below by 2 by induction as follows:
$x_1 \geq 2 \Rightarrow x_1-1 \geq 2-1=1$
$\Rightarrow x_2=1+\sqrt{x_1-1}\geq 1+1=2 $.
Let $x_k \geq 2 \Rightarrow x_k-1 \geq 2-1=1$
$\Rightarrow x_{k+1}=1+\sqrt{x_k-1}\geq 1+1=2 $.
How do I show that this is a decreasing sequence? 
Then my limit will be given by lim $x_{n+1}$=lim $(1+\sqrt{x_n-1})$
$\Rightarrow l=1+\sqrt {l-1}$
$\Rightarrow l-1=\sqrt {l-1}$
$\Rightarrow (l-1)^2=(l-1)$
$\Rightarrow (l-1)(l-2)=0$
Hence $l=1$ or $2$, but since it is bounded below by $2, l \neq 1$. So limit is $2$.
Please help with the decreasing proof!
 A: make a change of variable. let $y_n = x_n - 1.$ then $$y_{n+1} = \sqrt y_n,\quad y_1 \ge 1.$$
we have $$1 \le y_1 \implies 1 \le y_1 \le y_1^2=y_2 $$ and by induction, we have $$y_n \le y_{n+1}, n \ge 1.  $$
it can also  be seen from the fact that $\sqrt x$ has a unique fixed point $x = 1$ and from its graph.
A: Note that if $$x_{n+1}<x_n\implies 1+\sqrt {x_n-1}<x_n $$ $$\implies x_n-1<x_n^2-2x_n+1$$ $$\implies x_n^2-3x_n+2>0\implies (x_n-1)(x_n-2)>0$$ which is true since $x_n\geq 2$ for all $n$. Hence, the sequence is a decreasing sequence. Now, for a proof, you just reverse those implications and you can write: Since $x_n\geq 2$ for all $n$, $$(x_n-1)(x_n-2)>0 \implies x_n^2-3x_n+2>0$$ $$\implies x_n-1<x_n^2-2x_n+1\implies x_n-1<(x_n-1)^2$$ $$\implies \sqrt {x_n-1} <x_n-1\implies 1+\sqrt {x_n-1}<x_n \implies x_{n+1}<x_n$$
A: So to show it's decreasing, we need to show that $x_n > x_{n+1}$ (I will be using the conventional $n$). $\Leftarrow x_n > 1+\sqrt{x_n-1} \Leftarrow x_n-1 > \sqrt{x_n-1}$ which is true for all integers > 1. As we have shown above, $\lbrace x_n\rbrace \ge 2$, so this holds.
A: I would denote 
$$y_n:= x_n-1$$
Then you have $y_1 \geq 1$ and 
$$y_n=\sqrt{y_{n-1}}$$
You need to show that $y_n \geq 1$ and that $y_n \geq \sqrt{y_n}$, both of which are easy.
