Let $x_{n+1}=\sqrt{2+x_n}$ with $x_0=0$ and $y_{n+1}=\sqrt{2+y_n}$ with $y_0=2018$.
Do those two sequences have the same limit?
I think the answer is "Yes", but I am not really sure. Is my argument correct?
If $F(x)=\sqrt{2+x}$, then, $F'(x)<1$ for all $x>0$, so $F$ is contractive. Therefore, the above sequences converges by the contractive mapping theorem. The limit is the unique positive solution of the equation $s=\sqrt{2+s}$, which is $2$. Therefore, $\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}y_n=2$.
But then, we have the equality $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\cdots+\sqrt{2+\sqrt{2}}}}}}=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\cdots+\sqrt{2+\sqrt{2020}}}}}}$, which I think, is not correct.
hint
Let $F(x)=\sqrt{2+x}$.
$F$ is continuous at $[0,+\infty)$.
If $(x_n)$ converges to $L$, necessarily $$F(L)=L$$
or $$L^2-L-2=0$$ and
$$L\in \{-1,2\}$$
but $L\ge 0$, thus $$L=2$$
$F$ is increasing at $[0,+\infty)$
we just need to compare $x_0$ and $y_0$ with $L$.
we have. $$x_0\le x_n\le L\le y_n\le y_0$$