Question

Chứng tỏ

\(\dfrac{1}{3} + \dfrac{1}{3^2} +...+ \dfrac{1}{3^{99}}<1\)

Answer 1

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 \(\dfrac{1}{3}+\dfrac{1}{3^2}+..............+\dfrac{1}{3^{99}}< 1\rightarrowđpcm\)

Answer 2

Đặt:

\(S=\dfrac{1}{3}+\dfrac{1}{3^2}+.....+\dfrac{1}{3^{99}}\)

\(3S=3\left(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{99}}\right)\)

\(3S=1+\dfrac{1}{3}+.....+\dfrac{1}{3^{98}}\)

\(3S-S=\left(1+\dfrac{1}{3}+....+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{99}}\right)\)

\(2S=1-\dfrac{1}{3^{99}}\)

\(2S< 1\)

\(S< 1\rightarrowđpcm\)