Đặt \(A=\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{2009^3}\)
Ta CM công thức sau :
\(\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)
Thật vậy ta có : \(\left(n-1\right).n.\left(n+1\right)=\left(n-1\right)\left(n+1\right).n=\left(n^2-1\right).n=n^3-n< n^3\\ \Rightarrow\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)
Áp dụng công thức trên vào biểu thức A ; ta có :
\(A=\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{2009^3}\\ < \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2008.2009.2010}\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}-\dfrac{1}{2009.2010}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2009.2010}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.2009.2010}< \dfrac{1}{4}\)
Anh Tú xem xét bài e nhé !!
Làm thử, sai thì thôi đừng mắng :D
Đặt :
\(A=\dfrac{1}{2^3}+\dfrac{1}{3^3}+...............+\dfrac{1}{2009^3}\)
Ta thấy :
\(\dfrac{1}{2^3}< \dfrac{1}{1.2.3}\)
\(\dfrac{1}{3^3}< \dfrac{1}{2.3.4}\)
...........................
\(\dfrac{1}{2009^3}< \dfrac{1}{2008.2009.2010}\)
\(\Leftrightarrow A< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+............+\dfrac{1}{2008.2009.2010}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+........+\dfrac{1}{2008.2009}-\dfrac{1}{2009.2010}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}-\dfrac{1}{2009.2010}\)
Có j đó sai sai :-?
