Đặt `D=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{100^2}`
`<=>2D=2(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{100^2})`
`<=>2D=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{99^2}`
`<=>2D-D=[2(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{100^2})]-(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{100^2})`
`<=>D=1-\frac{1}{100}`
Do `D<1` hay `1-\frac{1}{100}<1`
`=>\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{100^2}<1` (đpcm)
