<s>[INST] Which is bigger, $3$ or $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$? Which is bigger, $3$ or  $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$ 
 I am not sure how to do it. I thought of squaring both sides and moving the $2$ to the other side, then squaring again etc. But I don't see a pattern in the left side.
 [/INST] Let $x$ be the given iterated radical.  Render
$x^2=2+\sqrt{3+\sqrt{4+\sqrt{5+\sqrt{6+...}}}}$
Now multiply $x$ by $\sqrt{3/2}$:
$(\sqrt{3/2})x=\sqrt{3+(3/2)\sqrt{4+\sqrt{5+\sqrt{6+...}}}}$
$=\sqrt{3+\sqrt{9+(9/4)\sqrt{5+\sqrt{6+...}}}}$
$=\sqrt{3+\sqrt{9+\sqrt{(405/16)+(81/16)\sqrt{6+...}}}}$
$=\sqrt{3+\sqrt{9+\sqrt{(405/16)+\sqrt{(19683/128)+...}}}}$
The second and subsequent radicands in $(\sqrt{3/2})x$ are larger than those in $x^2-2$ forcing
$x^2-2<(\sqrt{3/2})x$
$x^2-(\sqrt{3/2})x-2<0$
$x<\sqrt{3/8}+\sqrt{19/8}$ from the quadratic formula
The sum of two square roots is less than twice the square root of the average of the radicands:
$x<2\sqrt{11/8}, x^2<11/2<9, \color{blue}{x<3}$
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