Cho A = 1 + \(\frac{1}{2} + \frac{1}{3}... + \frac{1}{{{2^{100}} - 1}}\). Chứng minh 50 < A < 100

Lời giải:

A = 1 + \(\left( {\frac{1}{2} + \frac{1}{3}} \right) + \left( {\frac{1}{{{2^2}}} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}} \right) + \left( {\frac{1}{{{2^3}}} + \frac{1}{9} + ... + \frac{1}{{15}}} \right) + ... + \left( {\frac{1}{{{2^{99}}}} + ... + \frac{1}{{{2^{100}} - 1}}} \right)\)

Ta thấy \(\frac{1}{2} + \frac{1}{3} < \frac{1}{2} + \frac{1}{2} = \frac{1}{2}.2;\)

          \(\frac{1}{{{2^2}}} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} < \frac{1}{{{2^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{2^2}}} = \frac{1}{{{2^2}}}.4\)

…

Làm tương tự với các ngoặc còn lại, ta được:

A < \(1 + \frac{1}{2}.2 + \frac{1}{{{2^2}}}{.2^2} + \frac{1}{{{2^3}}}{.2^3} + ... + \frac{1}{{{2^{99}}}}{.2^{99}}\) = 1 + 1 + … + 1 = 100

Suy ra: A < 100.

Mặt khác:

A = 1 + \(\frac{1}{2} + \left( {\frac{1}{3} + \frac{1}{4}} \right) + \left( {\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}} \right) + ... + \left( {\frac{1}{{{2^{99}} + 1}} + ... + \frac{1}{{{2^{100}} - 1}} + \frac{1}{{{2^{100}}}}} \right) - \frac{1}{{{2^{100}}}}\)

Ta thấy: \(\frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{{{2^2}}}.2 = \frac{1}{2}\)

              \(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \frac{1}{{{2^3}}}.4\)\( = \frac{1}{2}\)

                     …

Làm tương tự với các ngoặc còn lại, ta được:

A > \(1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ... + \frac{1}{2} - \frac{1}{{{2^{100}}}}\) = 50 + 1 – \(\frac{1}{{{2^{100}}}}\) > 50

Vậy 50 < A < 100.