Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2011}}+\dfrac{1}{3^{2012}}\)
\(\Rightarrow\dfrac{1}{3}A=\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2012}}+\dfrac{1}{3^{2013}}\)
\(\Rightarrow A-\dfrac{1}{3}A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2011}}+\dfrac{1}{3^{2012}}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...-\dfrac{1}{3^{2012}}-\dfrac{1}{3^{2013}}\)\(\Rightarrow\dfrac{2}{3}A=\dfrac{1}{3}-\dfrac{1}{3^{2013}}< \dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}A< \dfrac{1}{3}\)
\(\Rightarrow A< \dfrac{1}{3}.\dfrac{3}{2}=\dfrac{1}{2}\)
Vậy \(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2011}}+\dfrac{1}{3^{2012}}< \dfrac{1}{2}\)
