Đặt :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+......................+\dfrac{1}{3^{99}}\)
\(\Leftrightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+................+\dfrac{1}{3^{98}}\)
\(\Leftrightarrow3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+..............+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+................+\dfrac{1}{3^{99}}\right)\)\(\Leftrightarrow2A=1-\dfrac{1}{3^{99}}< 1\)
\(\Leftrightarrow A< 1\)
Vậy \(\dfrac{1}{3}+\dfrac{1}{3^2}+..............+\dfrac{1}{3^{99}}< 1\rightarrowđpcm\)
Đặt:
\(S=\dfrac{1}{3}+\dfrac{1}{3^2}+.....+\dfrac{1}{3^{99}}\)
\(3S=3\left(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{99}}\right)\)
\(3S=1+\dfrac{1}{3}+.....+\dfrac{1}{3^{98}}\)
\(3S-S=\left(1+\dfrac{1}{3}+....+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{99}}\right)\)
\(2S=1-\dfrac{1}{3^{99}}\)
\(2S< 1\)
\(S< 1\rightarrowđpcm\)
