Ta có:
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{2019}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{2018}}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+....+\frac{1}{3^{2018}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{2019}}\right)\)
\(\Rightarrow2A=1+\frac{1}{3}+....+\frac{1}{3^{2018}}-\frac{1}{3}-\frac{1}{3^2}-.....-\frac{1}{3^{2019}}\)
\(\Rightarrow2A=1-\frac{1}{3^{2019}}\)
\(\Rightarrow\)\(A=\left(1-\frac{1}{3^{2019}}\right):2\)
\(\Rightarrow A=1:2-\frac{1}{3^{2019}}:2\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{3^{2019}}\)
\(\Rightarrow A< \frac{1}{2}\)
Vậy \(A< \frac{1}{2}\)
nhân cả 2 vế của A cho 3
3A=1+1/3+1/3^2+1/3^3+...+1/3^2018
3A-A=1+1/3+1/3^2+1/3^3+...+1/3^2018-(1/3+1/3^2+1/3^3+...+1/3^2018+1/3^2019)
2A=1-1/3^2019
2A<1
A<1/2
