Ta có: `\frac{1}{2^2}<\frac{1}{1.2}=\frac{1}{1} −\frac{1}{2}` ; `\frac{1}{3^2}<\frac{1}{2.3} =\frac{1}{2} −\frac{1}{3}` ; ...; `\frac{1}{100^2}<\frac{1}{99.100} =\frac{1}{99} −\frac{1}{100}`
\(\Rightarrow\ S < 5(1−\dfrac{1}{100})=5.\dfrac{99}{100}<5.1=5\)
\(\Rightarrow\ S<5\)
Lại có: \(\dfrac{1}{2^2} >\dfrac{1}{2.3} =\dfrac{1}{2} −\dfrac{1}{3} ; \dfrac{1}{3^2} >\dfrac{1}{3.4} =\dfrac{1}{3} −\dfrac{1}{4 }; \dfrac{1}{100^2} >\dfrac{1}{100.101} =\dfrac{1}{100}−\dfrac{1}{101}\)
`=>S>5(\frac{1}{2} −\frac{1}{101} )``=`\(5.\dfrac{101−2}{2.101}\)` =\frac{5.99}{2.101} ~2,45=>` S>2`
Vậy 2 < S < 5 => Đpcm
