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!

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.
