Question

Chứng minh rằng:

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

Answer 1

Đặt :

\(A=\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\)

\(\Leftrightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}\)

\(\Leftrightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\right)\)

\(\Leftrightarrow2A=1-\frac{1}{3^{99}}< 1\)

\(\Leftrightarrow A< \frac{1}{2}\left(đpcm\right)\)

Answer 2

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