Đặt: \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2019}{3^{2019}}\)
\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{2019}{3^{2018}}\)
\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}-\frac{2019}{3^{2019}}\)
Đặt: \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2017}}\)
\(\Rightarrow2B=1-\frac{1}{3^{2018}}\)
\(\Rightarrow B=\frac{1-\frac{1}{3^{2018}}}{2}\)
Thay vào \(2A\Rightarrow2A=1+\frac{\left(1-\frac{1}{3^{2018}}\right)}{2}-\frac{2019}{3^{2019}}\)
\(=1+\frac{1}{2}-\frac{1}{2.3^{2018}}-\frac{2019}{3^{2019}}< 1+\frac{1}{2}=\frac{3}{2}\)
\(\Rightarrow A< 0,75\left(đpcm\right)\)
