Ta có: \(n^3-n< n^3\forall n\)
mà: \(n^3-n=n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\)
Nên: \(\left(n-1\right)n\left(n+1\right)< n^3\Leftrightarrow\dfrac{1}{\left(n-1\right)n\left(n+1\right)}>\dfrac{1}{n^3}\)
Trở lại bài toán:
\(SV=\dfrac{1}{5^3}+\dfrac{1}{6^3}+\dfrac{1}{7^3}+...+\dfrac{1}{2004^3}< \dfrac{1}{\left(5-1\right).5.\left(5+1\right)}+\dfrac{1}{\left(6-1\right).6.\left(6+1\right)}+\dfrac{1}{\left(7-1\right).7.\left(7+1\right)}+...+\dfrac{1}{\left(2004-1\right).2004.\left(2004+1\right)}\)
\(SV< \dfrac{1}{4.5.6}+\dfrac{1}{5.6.7}+\dfrac{1}{6.7.8}+...+\dfrac{1}{2003.2004.2005}=\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}+\dfrac{1}{5.6}-\dfrac{1}{6.7}+\dfrac{1}{6.7}-\dfrac{1}{7.8}+...+\dfrac{1}{2003.2004}-\dfrac{1}{2004.2005}\right)=\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{2004.2005}\right)=\dfrac{1}{2.4.5}-\dfrac{1}{2.2004.2005}=\dfrac{1}{40}-\dfrac{1}{2.2004.2005}< \dfrac{1}{40}\left(đpcm\right)\)
