<s>[INST] Show that the sequence $a_0 = 1$, $a_{n+1 }= \sqrt{2+a_n}$ is monotonically increasing Given the sequence $a_0 = 1$, $a_{n+1}= \sqrt{2+a_n}$, how can I show that it is monotonically increasing?
I need it to show, that the sequence converges. I already proved boundedness but I can't figure out this last part.
 [/INST] Option 1: First, prove that $1 \le a_n < 2$ for all $n \in \mathbb{N}$. This can be done using induction. 
After that, prove $\sqrt{2+x} > x$ for $1 \le x < 2$, and thus, $a_{n+1} = \sqrt{2+a_n} > a_n$ for all $n \in \mathbb{N}$. 
Option 2: Define $\theta_n = \arccos\dfrac{a_n}{2}$. Then, $a_n = 2\cos\theta_n$ for each $n \in \mathbb{N}$. 
Hence, $2\cos\theta_{n+1} = a_{n+1} = \sqrt{2+a_n} = \sqrt{2+2\cos\theta_n} = 2\cos\tfrac{\theta_n}{2}$, and thus, $\theta_{n+1} = \dfrac{\theta_n}{2}$. 
Now it is pretty easy to prove that $\theta_n = \dfrac{\theta_0}{2^n}$ is monotonically decreasing to $0$, and thus, $a_n = 2\cos\theta_n$ monotonically increases to $2\cos 0 = 2$.
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