Đặt :
\(A=\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\)
\(\Leftrightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}\)
\(\Leftrightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\right)\)
\(\Leftrightarrow2A=1-\frac{1}{3^{99}}< 1\)
\(\Leftrightarrow A< \frac{1}{2}\left(đpcm\right)\)
Đặt \(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}\)
\(\Rightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(\Rightarrow2C=1-\frac{1}{3^{99}}< 1\)
=> C = (1 - 1/399)/2 < 1/2
Vậy 1/3 + 1/32 + 1/33 + ....+ 1/399 < 1/2