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.
Best Answer
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 $$a<b<3$$