\(\Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\\ \Rightarrow3A-A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2004}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{2003}}\\ \Rightarrow2A=1+\dfrac{1}{3^{2004}}\\ \Rightarrow A=\dfrac{1}{2}+\dfrac{1}{2\cdot3^{2004}}>\dfrac{1}{2}\)