Ta có:\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3A-A=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow2A=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}< \dfrac{1}{2}\)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(3A=3\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(3A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(3A-A=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(2A=1-\dfrac{1}{3^{99}}\)
\(A=\dfrac{1-\dfrac{1}{3^{99}}}{2}\\ \)
\(\text{Mà }1-\dfrac{1}{3^{99}}< 1\\ \Rightarrow\dfrac{1-\dfrac{1}{3^{99}}}{2}< \dfrac{1}{2}\\ \Rightarrow A< \dfrac{1}{2}\left(ĐPCM\right)\\ \)
Vậy \(A< \dfrac{1}{2}\)