Cho biểu thức \(B = \frac{1}{{{5^2}}} + \frac{1}{{{6^2}}} + \frac{1}{{{7^2}}} + ... + \frac{1}{{{{100}^2}}}\). Chứng tỏ rằng \(\frac{1}{6} < B < \frac{1}{4}\)

Lời giải:

Ta có: \(\frac{1}{{{5^2}}} > \frac{1}{{5.6}};\,\,\,\frac{1}{{{6^2}}} > \frac{1}{{6.7}};\,\,....\,;\frac{1}{{{{100}^2}}} > \frac{1}{{100.101}}\)

\(\begin{array}{l}\frac{1}{{{5^2}}} + \frac{1}{{{6^2}}} + ... + \frac{1}{{{{100}^2}}} > \frac{1}{{5.6}} + \frac{1}{{6.7}} + ... + \frac{1}{{100.101}}\\\frac{1}{{{5^2}}} + \frac{1}{{{6^2}}} + ... + \frac{1}{{{{100}^2}}} > \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7} + ... + \frac{1}{{100}} - \frac{1}{{101}}\\\frac{1}{{{5^2}}} + \frac{1}{{{6^2}}} + ... + \frac{1}{{{{100}^2}}} > \frac{1}{5} - \frac{1}{{101}}\\\frac{1}{{{5^2}}} + \frac{1}{{{6^2}}} + ... + \frac{1}{{{{100}^2}}} > \frac{{96}}{{505}} > \frac{1}{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\end{array}\)

Ta có: \(\frac{1}{{{5^2}}} < \frac{1}{{4.5}};\,\,\,\frac{1}{{{6^2}}} < \frac{1}{{5.6}};\,\,....\,;\frac{1}{{{{100}^2}}} < \frac{1}{{99.100}}\)

\(\begin{array}{l}\frac{1}{{{5^2}}} + \frac{1}{{{6^2}}} + ... + \frac{1}{{{{100}^2}}} < \frac{1}{{4.5}} + \frac{1}{{5.6}} + ... + \frac{1}{{99.100}}\\\frac{1}{{{5^2}}} + \frac{1}{{{6^2}}} + ... + \frac{1}{{{{100}^2}}} < \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + ... + \frac{1}{{99}} - \frac{1}{{100}}\\\frac{1}{{{5^2}}} + \frac{1}{{{6^2}}} + ... + \frac{1}{{{{100}^2}}} < \frac{1}{4} - \frac{1}{{100}}\\\frac{1}{{{5^2}}} + \frac{1}{{{6^2}}} + ... + \frac{1}{{{{100}^2}}} < \frac{6}{{25}} < \frac{1}{4}\,\,\,\,\,\,\,\,\,\,(2)\end{array}\)

Từ (1) và (2) suy ra \(\frac{1}{6} < B < \frac{1}{4}\) (đpcm)