Đề bài

Cho \(A = \frac{1}{3} + \frac{1}{{{3^3}}} + \frac{1}{{{3^5}}} + \frac{1}{{{3^7}}} + ... + \frac{1}{{{3^{2023}}}}\). Chứng minh rằng \(A < \frac{3}{8}\).

Phương pháp giải

Tính 9A.

Xét 9A – A.

Từ đó chứng minh được \(A < \frac{3}{8}\).

Lời giải của GV Loigiaihay.com

Ta có:

\(\begin{array}{c}9A = 9\left( {\frac{1}{3} + \frac{1}{{{3^3}}} + \frac{1}{{{3^5}}} + \frac{1}{{{3^7}}} + ... + \frac{1}{{{3^{2023}}}}} \right)\\ = 3 + \frac{1}{3} + \frac{1}{{{3^3}}} + \frac{1}{{{3^5}}} + ... + \frac{1}{{{3^{2021}}}}\end{array}\)

Xét \(9A - A = \left( {3 + \frac{1}{3} + \frac{1}{{{3^3}}} + \frac{1}{{{3^5}}} + ... + \frac{1}{{{3^{2021}}}}} \right) - \left( {\frac{1}{3} + \frac{1}{{{3^3}}} + \frac{1}{{{3^5}}} + \frac{1}{{{3^7}}} + ... + \frac{1}{{{3^{2023}}}}} \right)\)

\(\begin{array}{c}8A = 3 + \frac{1}{3} + \frac{1}{{{3^3}}} + \frac{1}{{{3^5}}} + ... + \frac{1}{{{3^{2021}}}} - \frac{1}{3} - \frac{1}{{{3^3}}} - \frac{1}{{{3^5}}} - \frac{1}{{{3^7}}} - ... - \frac{1}{{{3^{2023}}}}\\8A = 3 + \left( {\frac{1}{3} - \frac{1}{3}} \right) + \left( {\frac{1}{{{3^3}}} - \frac{1}{{{3^3}}}} \right) + ... + \left( {\frac{1}{{{3^{2021}}}} - \frac{1}{{{3^{2021}}}}} \right) - \frac{1}{{{3^{2023}}}}\end{array}\)

\(8A = 3 - \frac{1}{{{3^{2023}}}}\)

\(A = \frac{3}{8} - \frac{1}{{{{8.3}^{2023}}}}\)

Vì \(\frac{1}{{{{8.3}^{2023}}}} > 0\) nên \(A = \frac{3}{8} - \frac{1}{{{{8.3}^{2023}}}} < \frac{3}{8}\)

Vậy \(A < \frac{3}{8}\)