# How to solve this problem related to sequence?

The question is :

Let $\{x_n\}$ be a sequence such that $x_1 > \pi + \sqrt 2$ and $x_{n+1} = \pi + \sqrt {x_n - \pi}$.Then show that the sequence $\{x_n\}$ is convergent and it converges to $\pi + 1$.

How can I solve it?Please give me a hint (mainly about monotoneness).Then I will retry it.Thank you in advance.

Some useful facts:$$\pi+\sqrt{(\pi+1)-\pi}=\pi+1$$ If $x>\pi+1$, $$\pi+1<\pi+\sqrt{x-\pi}<x$$ If $\pi<x<\pi+1$, $$x<\pi+\sqrt{x-\pi}<\pi+1$$ Use these to prove that a limit exists. (Think: with each progressive iteration, the sequence gets closer and closer to $\pi+1$)
Once you know a limit exists, use the fact that $\lim_{n\to\infty}x_n=\lim_{n\to\infty}x_{n+1}$ to show that the limit is $\pi+1$
Put $y_n=x_n-\pi$. Is clear that $x_n$ converges iff $y_n$ too.
The problem is $y_1>\sqrt{2}$ and $y_{n+1}=\sqrt{y_n}$. Then, $y_n$ is decreasing and bounded by 1, $y_n$ converges to $l>0$. Now, $l=\sqrt{l}\to l=1$ and $x_n$ converges to $\pi+1$.
Comment: really, $\sqrt{2}$ is irrelevant, if we put $y_1=a\geq 1$, $y_n=a^{\frac{1}{2^{n-1}}}\to 1$