How to show that $\sqrt{1+\sqrt{2+\sqrt{3+\cdots\sqrt{2006}}}}<2$ $\sqrt{1+\sqrt{2+\sqrt{3+\cdots\sqrt{2006}}}}<2$.
I struggled on it, but I didn't find any pattern to solve it.
 A: Idea: You can unwrap like this:
$$\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}}<2$$
if
$$\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}<2^2-1$$
if
$$\sqrt{3+\cdots+\sqrt{2006}}<(2^2-1)^2-2$$
and so on, so we want to show
$$2006 < (((2^2-1)^2-2)^2-\cdots)^2-2005$$

might as well prove it by induction for all $n$ rather than just 2006, so we need to show that
$$n+1 < (((2^2-1)^2-2)^2-\cdots)^2-n$$
implies
$$n+2 < ((((2^2-1)^2-2)^2-\cdots)^2-n)^2-(n+1)$$
but thats just
$$2n+3 < ((((2^2-1)^2-2)^2-\cdots)^2-n)^2$$
which holds since
$$2n+3 < (n+1)^2$$
for all $n>1$.
A: $$\begin{aligned}\sqrt{1+\sqrt{2+\sqrt{3+\cdots \sqrt{n}}}}&<\sqrt{1+\sqrt{2+\sqrt{2^2+\cdots \sqrt{2^{2^{n-1}}}}}}\\&<\sqrt{1+\sqrt{2+\sqrt{2^2+\cdots }}}\\&=\sqrt{1+\sqrt{2}\cdot\sqrt{1+\sqrt{1+\cdots }}}\\&=\sqrt{1+\sqrt{2}\phi}\\&<2\end{aligned}$$

We can get a tighter bound for the limit by breaking the pattern a little further down the line:
$$\begin{aligned}\sqrt{1+\sqrt{2+\sqrt{3+\cdots \sqrt{n}}}}&<\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{2^2+\cdots}}}}\\&=\sqrt{1+\sqrt{2+\sqrt{3+2\sqrt{1+\cdots }}}}\\&=\sqrt{1+\sqrt{2+\sqrt{4+\sqrt{5}}}}\approx 1.7665398\end{aligned}$$
$$\sqrt{1+\sqrt{2+\sqrt{3+\cdots }}}\approx 1.7579327$$
A: I got a nice one! ($46^2=2116)$
$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+....+2016}}}}<\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{46^2}}}}=\sqrt{1+\sqrt{2+\sqrt{49}}}=\sqrt{1+\sqrt{9}}=2.$
