Đặt A=\(\frac{1}{3}+\frac{2}{3^2}+.....+\frac{2019}{3^{2019}}\)
3A=\(1+\frac{2}{3}+.....+\frac{2019}{3^{2018}}\)
3A - A = \(\left(1+\frac{2}{3}+...+\frac{2018}{3^{2017}}+\frac{2019}{3^{2018}}\right)\) -\(\left(\frac{1}{3}+....+\frac{2017}{3^{2017}}+\frac{2018}{3^{2018}}+\frac{2019}{3^{2019}}\right)\)
2A = \(1+\frac{1}{3}+...+\frac{1}{3^{2018}}-\frac{2019}{3^{2019}}\)
Đặt B=\(1+\frac{1}{3}+....+\frac{1}{3^{2018}}\)
3B =\(3+1+....+\frac{1}{3^{2017}}\)
3B - B=\(\left(3+1+....+\frac{1}{3^{2017}}\right)\)-\(\left(1+\frac{1}{3}+...+\frac{1}{3^{2018}}\right)\)
2B =\(3-\frac{1}{3^{2018}}\)
Ta có:2A= B - \(\frac{2019}{3^{2019}}\)
4A = 2B -\(\frac{2.2019}{3^{2019}}\)
4A=\(\left(3-\frac{1}{3^{2018}}\right)\)-\(\frac{2.2019}{3^{2019}}\)
A=\(\frac{3}{4}-\frac{1}{3^{2018}.4}-\frac{2019}{3^{2019}.2}\)<\(\frac{3}{4}\)=0,75
Suy ra :\(\frac{1}{3}+\frac{2}{3^2}+...+\frac{2019}{3^{2019}}\)< 0,75 (đpcm)