Question

Given $M=[\frac{1}{1^{\frac{2}{3}}}+\frac{1}{2^{\frac{2}{3}}}+\cdots +\frac{1}{{1000}^{\frac{2}{3}}}]$ (where [.] denotes greatest integer function), (M - 20) is equal to ___

Answer 1

${\int }_1^{1000}{x}^{\frac{-2}{3}}dx=3(x^{\frac{1}{3}}{)}_1^{1000}=3(10-1)=27and{\int }_1^{1000}{x}^{\frac{-2}{3}}dx<{\sum }_{n=1}^{999}{n}^{\frac{-2}{3}}\Rightarrow 27<{\sum }_{n=1}^{1000}{n}^{\frac{-2}{3}}-\frac{1}{(1000{)}^{\frac{2}{3}}}\therefore {\sum }_{n=1}^{1000}{n}^{\frac{-2}{3}}>27+\frac{1}{100}{\int }_1^{1000}{x}^{\frac{-2}{3}}dx>{\sum }_{n=2}^{1000}{n}^{\frac{-2}{3}}\Rightarrow 27>{\sum }_{n=1}^{1000}{n}^{\frac{-2}{3}}-1\Rightarrow {\sum }_{n=1}^{1000}{n}^{\frac{-2}{3}}<28\therefore M=[{\sum }_{n=1}^{1000}{n}^{\frac{-2}{3}}]=27$