\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\)
\(\Rightarrow\dfrac{9}{3}B=\dfrac{9}{3}\cdot\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2005}}\right)\)
\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\)
\(\Rightarrow3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+...+\left(1-\dfrac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=0+0+0+...+\left(1-\dfrac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=1-\dfrac{1}{3^{2005}}\)
\(\Rightarrow B=\dfrac{1-\dfrac{1}{3^{2005}}}{2}\)
\(\Rightarrow B=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3^{2005}}\right)\)
\(\Rightarrow B=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{2005}}\)
Mà: \(\dfrac{1}{2}-\dfrac{1}{2\cdot3^{2005}}< \dfrac{1}{2}\)
Nên: \(B< \dfrac{1}{2}\)
3B=1+1/3+...+1/3^2004
=>2B=1-1/3^2005
=>B=1/2-1/2*3^2005<1/2