<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] Intuitively, all those big numbers at the right are under so many square root signs they don't matter much.  I would think it "obvious" that $3$ is larger.  I made a spreadsheet (love copy down) and found the value to be about $2.0903$.  For comparison, I changed all the numbers after $9$ to $0$ and the value only changed in the sixth decimal place.  
Proving it without computer calculation is another matter.  One approach is just to unpack the square roots one by one.  We want to compare $9$ with ${2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}}}}$, which means comparing $7$ with $ \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}}}$, which means comparing $49$ with $3 + \sqrt{4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}}$, which means comparing $46$ with $ \sqrt{4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}}$, which means comparing $46^2$ with $4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}$ and now we can afford to take all the square root signs except the first off the right and still have the left larger- $46^2 \gt 4+\sqrt{5+6+7+8+\ldots 100}=4+\sqrt{5044}$.
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