` A = 1/3 + 1/(3^2) + 1/(3^3) + ... + 1/(3^{2020}) `
` → 3A = 1 + 1/3 + 1/(3^2) + ... 1/(3^{2019}) `
` → 3A - A = 1 + 1/3 + 1/(3^2) + ... 1/(3^{2019}) - 1/3 - 1/(3^2) - 1/(3^3) - ... - 1/(3^{2020}) `
` → 2A = 1 + (1/3 - 1/3) + (\frac{1}{3^2} - \frac{1}{3^2}) + ... + (\frac{1}{3^{2019}} - \frac{1}{3^{2019}}) - 1/(3^{2020}) `
` → 2A = 1 - 1/(3^{2020}) `
Đặt ` B = 1/2 `
Ta có:
` 2A = 1 - 1/(3^{2020}) `
` B = 1/2 → 2B = 1 `
Vì ` 1 - 1/(3^{2020}) < 1 `
` => 2A < 2B `
` => A < 1/2 `
Vậy ` A < 1/2 `
