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?
Best Answer
Your proof of growth is presented in a weird backward way. You could write
$$c_{n}\ge c_{n-1}\iff\sqrt{c_{n}+2}\ge\sqrt{c_{n-1}+2}\iff c_{n+1}\ge c_n.$$ You need a base case, which is missing. We indeed have $\sqrt{2+\sqrt2}\ge\sqrt2$.
You can also condense in a single, non-inductive proof of
$$0\le c_n\le2\implies c_n\le c_{n+1}\le2$$
or equivalently
$$0\le x\le2\implies x\le\sqrt{x+2}\le\sqrt{2+2}$$ because $$x^2-x-2\le0$$ for $x\in[-1,2]$.