Express sequence in closed type

Question

Given the sequence $a_n = \sqrt{2+a_{n-1}}$. Is there anyway to find a closed form for this sequence?

Thank you for your time.

Answer 1

If $a_0=0$ then, $$a_n=2\cos(\frac{\pi}{2^{n+1}})=\sqrt{2+\sqrt{2+\sqrt{2+....\sqrt{2}}}}$$

Where there are $n$ $2's$ on the right hand side.

This can easily be shown through the use of the trigonometric identity $$\cos(2x)=2\cos(x)^2-1$$

Taking the limit as $n$ tends to infinity shows that the nested radical clearly converges to two,

$$\lim_{n\to\infty} a_n=\lim_{n\to \infty}2\cos(\frac{\pi}{2^{n+1}})=2$$

Answer 2

We are given $a_n = \sqrt{2+a_{n-1}}$.

I will assime that $0 \le a_0 < 2$.

If $a_{n-1} < 2$, $a_n < \sqrt{2+2} = 2$, so all subsequent $a_n < 2$.

Since $a_1 = \sqrt{2+a_0} < 2$, all $a_n < 2$.

Let $d_n = 2-a_n$, so $a_n = 2-d_n$.

$2-d_n = \sqrt{2+(2-d_{n-1})} =\sqrt{4-d_{n-1}} $ or, since $1-x < \sqrt{1-x} < 1-x/2$ if $0 < x < 1$, $d_n = 2-\sqrt{4-d_{n-1}} = 2-2\sqrt{1-d_{n-1}/4} < 2-2(1-d_{n-1}/4) = d_{n-1}/2 $ so $d_n \to 0$ geometrically, so $a_n \to 2$.