Cho \(A = \frac{1}{{{2^2}}} + \frac{1}{{{4^2}}} + \frac{1}{{{6^2}}} + ... + \frac{1}{{{{100}^2}}}\). Chứng minh \(A < \frac{1}{2}\).

Ta có:  \(A = \frac{1}{{{2^2}}} + \frac{1}{{{4^2}}} + \frac{1}{{{6^2}}} + ... + \frac{1}{{{{100}^2}}}\)

               \( = \frac{1}{{{2^2}}}\left( {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + ... + \frac{1}{{{{50}^2}}}} \right)\).

Mặt khác: \(\frac{1}{{{2^2}}} = \frac{1}{{2.2}} < \frac{1}{{1.2}} = \frac{{2 - 1}}{{1.2}} = \frac{2}{{1.2}} - \frac{1}{{1.2}} = 1 - \frac{1}{2}\)

                  \(\frac{1}{{{3^2}}} = \frac{1}{{3.3}} < \frac{1}{{2.3}} = \frac{{3 - 2}}{{2.3}} = \frac{3}{{2.3}} - \frac{2}{{2.3}} = \frac{1}{2} - \frac{1}{3}\)

                  ………………..

                  \(\frac{1}{{{{50}^2}}} = \frac{1}{{50.50}} < \frac{1}{{49.50}} = \frac{{50 - 49}}{{49.50}} = \frac{{50}}{{49.50}} - \frac{{49}}{{49.50}} = \frac{1}{{49}} - \frac{1}{{50}}\)

Do đó  \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + ... + \frac{1}{{{{50}^2}}} < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{{49}} - \frac{1}{{50}}\)

Suy ra \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + ... + \frac{1}{{{{50}^2}}} < 1 - \frac{1}{{50}}\)

           \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + ... + \frac{1}{{{{50}^2}}} < \frac{{49}}{{50}} < \frac{{50}}{{50}} = 1\)

           \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + ... + \frac{1}{{{{50}^2}}} < 1\)

Từ đó ta có: \(1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + ... + \frac{1}{{{{50}^2}}} < 1 + 1 = 2\)

\[A = \frac{1}{{{2^2}}}\left( {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + ... + \frac{1}{{{{50}^2}}}} \right) < \frac{1}{4}.2 = \frac{1}{2}\].

Vậy \(A < \frac{1}{2}.\)