So sánh các số sau: \(A = \frac{1}{{1000}};B = {\left( {\frac{{99}}{{1000}}} \right)^3};C = {\left( {\frac{1}{{{{11}^2}}} + \frac{1}{{{{12}^2}}} + ... + \frac{1}{{{{1000}^2}}}} \right)^3}\).
Nếu \(n > 0\) thì \(a > b > 0 \Rightarrow {a^n} > {b^n}\)
Do \(0 < \frac{{99}}{{1000}} < \frac{{100}}{{1000}} = \frac{1}{{10}} < 1 \Rightarrow {\left( {\frac{{99}}{{1000}}} \right)^3} < {\left( {\frac{1}{{10}}} \right)^3} \Rightarrow B < A\left( 1 \right)\)
Với \(n \in \mathbb{N}*,n > 1\) ta có: \(\frac{1}{{n - 1}} - \frac{1}{n} = \frac{{n - \left( {n - 1} \right)}}{{n\left( {n - 1} \right)}} = \frac{1}{{n\left( {n - 1} \right)}} > \frac{1}{{{n^2}}} \Rightarrow \frac{1}{{{n^2}}} < \frac{1}{{n - 1}} - \frac{1}{n}\)
Do đó, \(\frac{1}{{{{11}^2}}} < \frac{1}{{10}} - \frac{1}{{11}};\frac{1}{{{{12}^2}}} < \frac{1}{{11}} - \frac{1}{{12}};...;\frac{1}{{{{1000}^2}}} < \frac{1}{{999}} - \frac{1}{{1000}}\)
\( \Rightarrow \frac{1}{{{{11}^2}}} + \frac{1}{{{{12}^2}}} + ... + \frac{1}{{{{1000}^2}}} < \frac{1}{{10}} - \frac{1}{{11}} + \frac{1}{{11}} - \frac{1}{{12}} + ... + \frac{1}{{999}} - \frac{1}{{1000}} = \frac{1}{{10}} - \frac{1}{{1000}} = \frac{{99}}{{1000}}\)
\( \Rightarrow {\left( {\frac{1}{{{{11}^2}}} + \frac{1}{{{{12}^2}}} + ... + \frac{1}{{{{1000}^2}}}} \right)^3} < {\left( {\frac{{99}}{{1000}}} \right)^3} \Rightarrow C < B\left( 2 \right)\)
Từ (1) và (2) ta có: \(C < B < A\).
