\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\)
\(3B=3.\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\right)\)
\(3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\)
\(3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\right)\)
\(2B=1-\dfrac{1}{3^{2005}}\)\(< 1\)
\(\Rightarrow B< \dfrac{1}{2}\) ( đpcm )
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