Tham khảo
`A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}`
`⇒3A=1+\frac{1}{3}+....+\frac{1}{3^{2013}}`
`⇒3A-A=1+\frac{1}{3}+....+\frac{1}{3^{2013}}-(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}})`
`⇒2A=1-\frac{1}{3^{2014}}<1`
`⇒A<\frac{1}{2}<\frac{3}{2}`
Vậy `A<\frac{3}{2}`
`\text{©CBT}`