Cho \(M = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{3^4}}} + \frac{1}{{{3^6}}} + ... + \frac{1}{{{3^{802}}}}\). Chứng minh rằng \(M < \frac{3}{8}\).
Đặt \(A = \frac{1}{{{3^2}}} + \frac{1}{{{3^4}}} + \frac{1}{{{3^6}}} + ... + \frac{1}{{{3^{802}}}}\)
Nhân hai vế của \(A\) với \({3^2}\).
Lấy \({3^2}A - A\), so sánh với 1 để chứng minh \(A < \frac{1}{8}\).
Từ đó chứng minh \(M = \frac{1}{{{2^2}}} + A < \frac{3}{8}\)
Đặt \(A = \frac{1}{{{3^2}}} + \frac{1}{{{3^4}}} + \frac{1}{{{3^6}}} + ... + \frac{1}{{{3^{802}}}}\)
Ta có: \({3^2}.A = {3^2}.\left( {\frac{1}{{{3^2}}} + \frac{1}{{{3^4}}} + \frac{1}{{{3^6}}} + ... + \frac{1}{{{3^{802}}}}} \right)\)
\(9A = 1 + \frac{1}{{{3^2}}} + \frac{1}{{{3^4}}} + \frac{1}{{{3^6}}} + ... + \frac{1}{{{3^{800}}}}\)
Suy ra
\(9A - A = \left( {1 + \frac{1}{{{3^2}}} + \frac{1}{{{3^4}}} + \frac{1}{{{3^6}}} + ... + \frac{1}{{{3^{800}}}}} \right) - \left( {\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{3^4}}} + \frac{1}{{{3^6}}} + ... + \frac{1}{{{3^{802}}}}} \right)\)
\(8A = 1 - \frac{1}{{{3^{802}}}}\)
Vì \(1 - \frac{1}{{{3^{802}}}} < 1\) nên \(8A < 1\), suy ra \(A < \frac{1}{8}\).
Mà \(M = \frac{1}{{{2^2}}} + A < \frac{1}{4} + \frac{1}{8} = \frac{3}{8}\) nên \(M < \frac{3}{8}\).
Vậy \(M < \frac{3}{8}\).