Show that the sequence $a_0 = 1$, $a_{n+1 }= \sqrt{2+a_n}$ is monotonically increasing

Question

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.

Answer 1

We prove the result by induction. We can verify directly that $a_1\gt a_0$. For the induction step, suppose that for some specific $k$ we know that $a_{k+1}\gt a_{k}$. We will show that $a_{k+2}\gt a_{k+1}$. This is easy, for $$a_{k+2}=\sqrt{2+a_{k+1}}\gt \sqrt{2+a_k}=a_{k+1}.$$

Answer 2

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 $2\cos 0 = 2$.

Answer 3

Note that \begin{align*} a_{n+1} \ge a_n & \iff \sqrt{2+ a_n} \ge a_n \\ & \iff a_n^2 - a_n -2 = (a_n +1 ) (a_n-2) \le 0\end{align*}

So we apply induction. Assume $-1 \le a_k \le 2$ and show $-1 \le \sqrt{2 + a_k}\le 2$. (Of course we have to check $-1 \le a_1 = 1 \le 2$ )