Cho tổng gồm 2019 số hạng S=1/4+2/4^2 + 3/4^3 + 4/4^4 +...+ 2019/4^2019. Chứng minh rằng :A < 1/2

cho tổng gồm 2019 số hạng S= $\frac{1}{4} + \frac{2}{4^2} + \frac{3}{4^3} + \frac{4}{4^4} + . . . + \frac{2019}{4^2019}$ .CMR :A< $\frac{1}{2}$