Chuyên đề: Lũy thừa của một số hữu tỉ (phần 4)

Dạng 4: Chứng minh

*) Chứng minh bất đẳng thức:

Bài 1: Chứng minh rằng:

\(\begin{array}{l}a)\,H = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{2007}^2}}} + \frac{1}{{{{2008}^2}}}\,\, < \,\,1\\b)\,K = \frac{1}{{{2^2}}} + \frac{1}{{{4^2}}} + \frac{1}{{{6^2}}} + \frac{1}{{{8^2}}} + \frac{1}{{{{10}^2}}} + \frac{1}{{{{12}^2}}} + \frac{1}{{{{14}^2}}}\,\, < \,\,\frac{1}{2}\end{array}\)

Giải:

Lưu ý: \(\frac{1}{{n\left( {n + 1} \right)}} = \frac{1}{n} - \frac{1}{{n + 1}}\,\,\,\,\left( {n \in {N^*}} \right)\)

Ta có: \(\frac{1}{{{2^2}}}\,\, < \,\,\frac{1}{{1.2}}\,;\,\,\frac{1}{{{3^2}}}\,\, < \,\,\frac{1}{{2.3}};\,\,\frac{1}{{{4^2}}}\,\, < \,\,\frac{1}{{3.4}};\,.....;\,\,\frac{1}{{{{2008}^2}}}\,\, < \,\,\frac{1}{{2007.2008}}\)

\( \Rightarrow H = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{2007}^2}}} + \frac{1}{{{{2008}^2}}}\,\,\, < \,\,\,\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + \,... + \frac{1}{{2006.2007}} + \frac{1}{{2007.2008}}\)

Mà \(\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + \,... + \frac{1}{{2006.2007}} + \frac{1}{{2007.2008}} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{2006}} - \frac{1}{{2007}} + \frac{1}{{2007}} - \frac{1}{{2008}} = 1 - \frac{1}{{2008}}\,\,\, < \,\,1\)

\( \Rightarrow H\,\, < \,\,1\)