+ Từ bài toán tổng quát
(n-1).n.(n+1)=n3 - n => n3 = (n-1).n.(n+1) + n
\(\Rightarrow\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2006^3}=\)
\(=\frac{1}{1.2.3+2}+\frac{1}{2.3.4+3}+\frac{1}{3.4.5+4}+\frac{1}{2005.2006.2007-2006}=A\)
\(\Rightarrow A< \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2005.2006.2007}=B\)
\(\Rightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2005.2006.2007}\)
\(2B=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2007-2005}{2005.2006.2007}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2005.2006}-\frac{1}{2006.2007}\)
\(2B=\frac{1}{2}-\frac{1}{2006.2007}\Rightarrow B=\frac{1}{4}-\frac{1}{2.2006.2007}< \frac{1}{4}\)
\(\Rightarrow A< \frac{1}{4}\)
