Bài 6:
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2020}}+\dfrac{1}{3^{2021}}\)
=>\(3B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2019}}+\dfrac{1}{3^{2020}}\)
=>\(2B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2019}}+\dfrac{1}{3^{2020}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{2020}}-\dfrac{1}{3^{2021}}\)
=>\(2B=1-\dfrac{1}{3^{2021}}\)
=>\(B=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{2021}}< \dfrac{1}{2}\)