Cho \[A = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} +  \cdot  \cdot  \cdot  \cdot  \cdot  + \frac{1}{{{{2012}^2}}} + \frac{1}{{{{2013}^2}}}\]. Chứng tỏ \[A < 1\].

Ta có \[A = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} +  \cdot  \cdot  \cdot  \cdot  \cdot  + \frac{1}{{{{2012}^2}}} + \frac{1}{{{{2013}^2}}}\].

Đặt \(B = \frac{1}{{1.2}} + \frac{1}{{2.3}} + ....... + \,\frac{1}{{2012.2013}}\).

Ta có vì \[2 > 1\] nên \[2\,\,.\,\,2 > 1\,\,.\,\,2\].

Suy ra \(\frac{1}{{{2^2}}} = \frac{1}{{2.2}} < \frac{1}{{1.2}}\);

Tương tự:

\(\frac{1}{{{3^2}}} = \frac{1}{{3.3}} < \frac{1}{{2.3}}\);

….

\(\frac{1}{{{{2012}^2}}} = \frac{1}{{2012.2012}} < \frac{1}{{2011.2012}}\);

\(\frac{1}{{{{2013}^2}}} = \frac{1}{{2013.2013}} < \frac{1}{{2012.2013}}\).

Do đó \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} +  \cdot  \cdot  \cdot  \cdot  \cdot  + \frac{1}{{{{2012}^2}}} + \frac{1}{{{{2013}^2}}} < \frac{1}{{1.2}} + \frac{1}{{2.3}} + ....... + \frac{1}{{2011.2012}} + \,\frac{1}{{2012.2013}}\).

Suy ra \[A < \;B\].

Mà \(B = \frac{1}{{1.2}} + \frac{1}{{2.3}} + ....... + \,\frac{1}{{2012.2013}}\)

\( = \frac{{2 - 1}}{{1.2}} + \frac{{3 - 2}}{{2.3}} + ... + \frac{{2012 - 2011}}{{2011.2012}} + \frac{{2013 - 2012}}{{2012.2013}}\)

\( = \frac{2}{{1.2}} - \frac{1}{{1.2}} + \frac{3}{{2.3}} - \frac{2}{{2.3}} + ... + \frac{{2012}}{{2011.2012}} - \frac{{2011}}{{2011.2012}} + \frac{{2013}}{{2012.2013}} - \frac{{2012}}{{2012.2013}}\)

\[ = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{{2012}} - \frac{1}{{2013}}\]\( = 1 - \frac{1}{{2013}} < 1\).

Do đó \[B < 1\] nên \[A < B < 1\].

Vậy \[A < 1\].