<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] Actually, there is a rather simple method to do this:
We can rewrite 
$$a=\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$$
into 
$$b=\sqrt{2 + \sqrt{3 + \sqrt{100 + \sqrt{100 + \sqrt{100+\cdots}}}}}$$
And we can clearly see that
$$b>a$$
There is a classic method to evaluate continued root. Let 
$$x=\sqrt{100 + \sqrt{100 + \sqrt{100+\cdots}}}$$
Then 
$$x=\sqrt{100+x}$$
$$x^2=100+x$$
$$x=\frac{1+\sqrt{401}}{2}$$
Obviously,
$$20<\sqrt{401}<21$$
We see that
$$x<\frac{1+21}{2}=11$$
Then we have 
$$b<\sqrt{2 + \sqrt{3+11}}$$
$$b<\sqrt{2 + \sqrt{14}}<\sqrt{6}<3$$
So
$$a<b<3$$
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