Đặt VT=A
Ta có: [tex]\frac{A}{3}=\frac{1}{3^2}+...+\frac{1}{3^{100}}=>A-\frac{A}{3}=(\frac{1}{3}+...+\frac{1}{3^{99}})-(\frac{1}{3^2}+...+\frac{1}{3^{100}})<=>\frac{2A}{3}=\frac{1}{3}-\frac{1}{3^{100}}<=>\frac{2A}{3}<\frac{1}{3}<=>A<\frac{1}{2}[/tex]