I know this question has been asked many times, but I need a specific part of it.
When we get to the limit part, I had it written like this: $\lim_{n \to +\infty} a_{n+1} = \sqrt{2+\lim_{n \to +\infty} a_n}$
What is the reason I am allowed to put the limit under the square root? Why am I just allowed to put it there? I know it's true, but I am not sure why. I need to justify that in order to receive credit. My instructor said it is a "special word." I have no idea what the word is.
On
We have a few claims:
Claim: The sequence $\{ a_{n}\}_{n = 1}^\infty$ is bounded above by $2$.
Proof. The proof can be done by induction. I'll just write the induction step $k + 1$. $$ a_{k + 2}^2 = 2 + a_{k + 1} < 2 + 2 \implies a_{k + 2} < 2 $$ The rest is as goes with the formal writing.
Lemma: $\{ a_n\} = \left\{ 2 \cos \frac{\pi}{2^{n + 1}} \right\} $
Proof. Similar by induction. Assume $$ a_k = 2\cos \left( \frac{\pi}{2^{k + 1}} \right) $$ Then $$ a_{k +1}^2 = 2 + 2 \cos\left( \frac{\pi}{2^{k + 1}} \right) = 2^2 \cos^2 \left( \frac{\pi}{2^{k + 2}} \right) $$ $$ \implies a_{k + 1} = 2 \cos \left( \frac{\pi}{2^{k + 2}} \right) $$ Hence inductively we are done.
Now, we have a specific closed form. Then we have that $\lim\limits_{n \to \infty} a_n$ exists, as $\cos \left( \dfrac{\pi}{2^{n + 1}} \right) \to \cos 0 = 1 $, as $\cos$ is a continuous function on $\left[0, \dfrac{\pi}{2} \right]$.
Thus finally we have that $$ \lim_{n \to \infty} a_n = \lim_{n \to \infty} 2\cos \left( \frac{\pi}{2^{n + 1}} \right) = 2 $$
Edit: I have tried to find a closed-form.
IF the limit exists, it will be a fixed point of the function $\sqrt{2+x}$, in other words a solution to the equation $$x=\sqrt{2+x}$$ Here's why: $$\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\sqrt{2+a_n}$$ Given two functions $f$ and $g$, as long as $f$ is continuous and $\lim_{x\to x_0}g(x)$ exists, then $$\lim_{x\to x_0}f(g(x))=f\left(\lim_{x\to x_0}g(x)\right)$$ This can be shown fairly routinely with the $\epsilon ,\delta$ definition of the limit.
Since $\sqrt{2+x}$ is continuous on its domain, $$\lim_{n\to\infty}\sqrt{2+a_n}=\sqrt{2+\lim_{n\to\infty}a_n}=\lim_{n\to\infty}a_{n+1}$$ Since $x:=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}a_n$, the initial statement follows.