Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...-\frac{2014}{3^{2014}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-....-\frac{2014}{3^{2013}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-...-\frac{2014}{3^{2013}}\right)+\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-....-\frac{2014}{3^{2014}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{2013}}-\frac{2014}{3^{2014}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{2013}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{2013}}\)
3B = \(3-1+\frac{1}{3}-...-\frac{1}{3^{2012}}\)
3B + B = \(\left(3-1+\frac{1}{3}-...-\frac{1}{3^{2012}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{2013}}\right)\)
4B = \(3-\frac{1}{3^{2013}}\)
=> 4B < 3 => B < \(\frac{3}{4}\)(2)
Từ (1)(2) => 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)<\(\frac{1}{5}\)(dpcm)
