\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2015}}\)
\(\Rightarrow3B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2014}}\)
\(\Rightarrow3B-B=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{2014}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2015}}\right)\)
\(\Rightarrow2B=1-\dfrac{1}{3^{2015}}\)
\(\Rightarrow B=\left(1-\dfrac{1}{3^{2015}}\right).\dfrac{1}{2}=\dfrac{1}{2}-\dfrac{1}{3^{2015}.2}< \dfrac{1}{2}\)
\(\Rightarrowđpcm\)
Vậy...
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2015}}\)
\(\Rightarrow\dfrac{1}{3}B=\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2006}}\)
\(\Rightarrow B-\dfrac{1}{3}B=\dfrac{1}{3}-\dfrac{1}{3^{2006}}\)
\(\Rightarrow\dfrac{2}{3}B=\dfrac{1}{3}-\dfrac{1}{3^{2006}}\)
\(\Rightarrow B=\dfrac{1}{2}\left(1-\dfrac{1}{3^{2005}}\right)< \dfrac{1}{2}\)
\(\RightarrowĐpcm\)
