Question

[Minz Ank]

Chứng tỏ rằng: 1/3^3 + 1/5^3 + 1/7^3 +..+ 1/2021^3 < 1/12

Answer 1

[Nguyễn Hoàng Minh]

Ta có \(\dfrac{1}{3^3}< \dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\)

\(\dfrac{1}{5^3}< \dfrac{1}{4.5.6}=\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}\right)\\ ...\\ \dfrac{1}{2021^3}< \dfrac{1}{2020.2021.2022}=\dfrac{1}{2}\left(\dfrac{1}{2020.2021}-\dfrac{1}{2021.2022}\right)\)

Cộng VTV ta được

\(VT< \dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{2021.2022}\right)=\dfrac{1}{12}-\dfrac{1}{2\left(2021.2022\right)}< \dfrac{1}{12}\)

Answer 2

[Vinh 2k8]

\(n^3=n.n^2>n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\)

\(\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)

\(\dfrac{1}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{2}.\dfrac{n+1-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)\)

\(\dfrac{1}{3^3}+\dfrac{1}{5^3}+.......+\dfrac{1}{2009^3}< \dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+.....\dfrac{1}{2008.2009.2010}=\dfrac{1}{2}\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+.........+\dfrac{1}{2008.2009}-\dfrac{1}{2009.2010}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{2009.2010}\right)\)

\(=\dfrac{1}{2}\)