So sánh giá trị của biểu thức A = 1/ 2 − 1 /2^2 + 1/ 2^3 − 1 /2^4 + . . . + 1/ 2^99 − 1/ 2^100 với 1/3 .

Hướng dẫn giải:

Ta có: \(A = \frac{1}{2} - \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} - \frac{1}{{{2^4}}} + ... + \frac{1}{{{2^{99}}}} - \frac{1}{{{2^{100}}}}\)

Suy ra \(2A = 2 \cdot \left( {\frac{1}{2} - \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} - \frac{1}{{{2^4}}} + ... + \frac{1}{{{2^{99}}}} - \frac{1}{{{2^{100}}}}} \right)\)

                 \( = 1 - \frac{1}{2} + \frac{1}{{{2^2}}} - \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{98}}}} - \frac{1}{{{2^{99}}}}.\)

Do đó \[2A + A = \left( {1 - \frac{1}{2} + \frac{1}{{{2^2}}} - \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{98}}}} - \frac{1}{{{2^{99}}}}} \right) + \left( {\frac{1}{2} - \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} - \frac{1}{{{2^4}}} + ... + \frac{1}{{{2^{99}}}} - \frac{1}{{{2^{100}}}}} \right)\]

Suy ra \[3A = 1 - \frac{1}{{{2^{100}}}}\]

Nên \[A = \frac{1}{3} - \frac{1}{{3 \cdot {2^{100}}}}\]

Vì \[\frac{1}{{3 \cdot {2^{100}}}} > 0\] nên \(\frac{1}{3} - \frac{1}{{3 \cdot {2^{100}}}} < \frac{1}{3}.\)

Vậy \(A < \frac{1}{3}.\)