Câu hỏi:

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Cho A=122+132+      +120122+120132A = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} +  \cdot  \cdot  \cdot  \cdot  \cdot  + \frac{1}{{{{2012}^2}}} + \frac{1}{{{{2013}^2}}}. Chứng tỏ A<1A < 1.

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Lời giải

Hướng dẫn giải:

Ta có A=122+132+      +120122+120132A = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} +  \cdot  \cdot  \cdot  \cdot  \cdot  + \frac{1}{{{{2012}^2}}} + \frac{1}{{{{2013}^2}}}.

Đặt B=11.2+12.3+.......+12012.2013B = \frac{1}{{1.2}} + \frac{1}{{2.3}} + ....... + \,\frac{1}{{2012.2013}}.

Ta có vì 2>12 > 1 nên 2  .  2>1  .  22\,\,.\,\,2 > 1\,\,.\,\,2.

Suy ra 122=12.2<11.2\frac{1}{{{2^2}}} = \frac{1}{{2.2}} < \frac{1}{{1.2}};

Tương tự:

132=13.3<12.3\frac{1}{{{3^2}}} = \frac{1}{{3.3}} < \frac{1}{{2.3}};

….

120122=12012.2012<12011.2012\frac{1}{{{{2012}^2}}} = \frac{1}{{2012.2012}} < \frac{1}{{2011.2012}};

120132=12013.2013<12012.2013\frac{1}{{{{2013}^2}}} = \frac{1}{{2013.2013}} < \frac{1}{{2012.2013}}.

Do đó 122+132+      +120122+120132<11.2+12.3+.......+12011.2012+12012.2013\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<  BA < \;B.

B=11.2+12.3+.......+12012.2013B = \frac{1}{{1.2}} + \frac{1}{{2.3}} + ....... + \,\frac{1}{{2012.2013}}

=211.2+322.3+...+201220112011.2012+201320122012.2013 = \frac{{2 - 1}}{{1.2}} + \frac{{3 - 2}}{{2.3}} + ... + \frac{{2012 - 2011}}{{2011.2012}} + \frac{{2013 - 2012}}{{2012.2013}}

=21.211.2+32.322.3+...+20122011.201220112011.2012+20132012.201320122012.2013 = \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}}

=112+1213+...+1201212013 = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{{2012}} - \frac{1}{{2013}}=112013<1 = 1 - \frac{1}{{2013}} < 1.

Do đó B<1B < 1 nên A<B<1A < B < 1.

Vậy A<1A < 1.

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