A recursive sequence $c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$ We have a recursive sequence $$c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$$ with $n$ square roots.
We can obtain a recursive formula: $$c_n =\sqrt{2+c_{n-1}}\\
c_{n+1}=\sqrt{2+c_n}$$
Now we show that the sequence is increasing:
$$c_{n+1} \geq c_n \rightarrow\text{  This is what we need to prove}$$
$$ 
\sqrt{2+c_n} \geq \sqrt{2+c_{n-1}}\\
2+c_n \geq 2+c_{n-1}\\
c_n \geq c_{n-1} \implies c_{n+1} \geq c_n 
$$
Let's assume that a limit $c$ exists:
$$c = \sqrt{2+c}$$
$$c^2-c-2 = 0 \iff (c-2)(c-1)=0$$
So if the limit exists it is either 2 or 1. We know that it is not 1 since $c_1 > 1$ and the sequence is increasing.
$$c_n \leq 2\\
c_1 \text{ holds}\\$$
Now let's see for $n\rightarrow n+1$:
$$
c_{n+1} \leq 2\\
\sqrt{2+c_{n}} \leq \sqrt{2+2}\leq 2$$
Is the proof that the sequence is increasing  and is bounded sufficient?

Assuming the nested radical is infinite we have $x=\sqrt{2+\sqrt{2+...+\sqrt{2}}}\Rightarrow x^2-2=x \iff x^2-x-2=0\Rightarrow x=2$
