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2016 preRMO p12, 1 + 1/\sqrt2+1/\sqrt3+..+ 1/\sqrt{99}+ 1/\sqrt{100}

Source:

August 9, 2019
Sumradicalfloor functioninequalitiesalgebra

Problem Statement

Let S=1+12+13+14+...+199+1100S = 1 + \frac{1}{\sqrt2}+ \frac{1}{\sqrt3}+\frac{1}{\sqrt4}+...+ \frac{1}{\sqrt{99}}+ \frac{1}{\sqrt{100}} . Find [S][S].
You may use the fact that n<12(n+n+1)<n+1\sqrt{n} < \frac12 (\sqrt{n} +\sqrt{n+1}) <\sqrt{n+1} for all integers n1n \ge 1.