Cho \(S = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + .... + \frac{1}{{{{99}^2}}}.\) Chứng mính rằng \(\frac{{49}}{{100}} < S < 1.\)

Hướng dẫn giải

Ta có: \[\frac{1}{{{2^2}}} = \frac{1}{{2 \cdot 2}} < \frac{1}{{1 \cdot 2}};\]

 \[\frac{1}{{{3^2}}} = \frac{1}{{3 \cdot 3}} < \frac{1}{{2 \cdot 3}};\]

 \[\frac{1}{{{4^2}}} = \frac{1}{{4 \cdot 4}} < \frac{1}{{3 \cdot 4}};\]

 \[........................\]

\[\frac{1}{{{{99}^2}}} = \frac{1}{{99 \cdot 99}} < \frac{1}{{98 \cdot 99}}.\]

Suy ra \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{99}^2}}} < \frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} + ... + \frac{1}{{98 \cdot 99}}\)

\(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{99}^2}}} < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{98}} - \frac{1}{{99}}\)

\(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{99}^2}}} < 1 - \frac{1}{{99}}\)

\(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{99}^2}}} < \frac{{98}}{{99}} < 1\)

Do đó \[S = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{99}^2}}} < 1.\,\,\,\left( 1 \right)\]

Ta có: \(\frac{1}{{{2^2}}} = \frac{1}{{2 \cdot 2}} > \frac{1}{{2 \cdot 3}};\)

 \(\frac{1}{{{3^2}}} = \frac{1}{{3 \cdot 3}} > \frac{1}{{3 \cdot 4}};\)

 \(\frac{1}{{{4^2}}} = \frac{1}{{4 \cdot 4}} > \frac{1}{{4 \cdot 5}};\)

 \(........................\)

 \(\frac{1}{{{{99}^2}}} = \frac{1}{{99 \cdot 99}} > \frac{1}{{99 \cdot 100}}.\)

Suy ra \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{99}^2}}} > \frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + ... + \frac{1}{{99 \cdot 100}}\)

 \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{99}^2}}} > \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + .... + \frac{1}{{99}} - \frac{1}{{100}}\)

 \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{99}^2}}} > \frac{1}{2} - \frac{1}{{100}}\)

 \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{99}^2}}} > \frac{{49}}{{100}}{\rm{    }}\)

Do đó \(S = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{99}^2}}} > \frac{{49}}{{100}}.\,\,\,\left( 2 \right)\)

Từ (1) và (2) suy ra  \(\frac{{49}}{{100}} < S < 1.\)