n ≥ 2 n ∈ N 1/2^3 + 1/3^3 + 1/4^3 + …….. + 1/n^3 < 1/4 13/08/2021 Bởi Madelyn n ≥ 2 n ∈ N 1/2^3 + 1/3^3 + 1/4^3 + …….. + 1/n^3 < 1/4
`=>` Tặng bạn Ta có : $\(A=\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+……+\dfrac{1}{n^3}\)$ $\(2A=\dfrac{2}{2^3}+\dfrac{2}{3^3}+\dfrac{2}{4^3}+…….+\dfrac{2}{n^3}\)$ Vì : $\(\dfrac{2}{2^3}< \dfrac{2}{1.2.3}\)$ $\(\dfrac{2}{3^3}< \dfrac{1}{2.3.4}\)$ …………………………… $\(\dfrac{2}{n^3}< \dfrac{2}{\left(n-1\right)n\left(n+1\right)}\)$ $\(\Rightarrow2A< \dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+……………….+\dfrac{2}{\left(n-1\right)n\left(n+1\right)}\)$ $\(2A< \dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+…………..+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\)$ $\(2A< \dfrac{1}{1.2}-\dfrac{1}{n\left(n+1\right)}\)$ $\(\Rightarrow A< \left(\dfrac{1}{1.2}-\dfrac{1}{n\left(n+1\right)}\right):2\)$ $\(A< \dfrac{1}{4}-\dfrac{1}{2n\left(n+1\right)}\)$ $\(\Rightarrow A< \dfrac{1}{4}\) \(\rightarrowđpcm\)$ Bình luận
`=>` Tặng bạn
Ta có :
$\(A=\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+……+\dfrac{1}{n^3}\)$
$\(2A=\dfrac{2}{2^3}+\dfrac{2}{3^3}+\dfrac{2}{4^3}+…….+\dfrac{2}{n^3}\)$
Vì :
$\(\dfrac{2}{2^3}< \dfrac{2}{1.2.3}\)$
$\(\dfrac{2}{3^3}< \dfrac{1}{2.3.4}\)$
……………………………
$\(\dfrac{2}{n^3}< \dfrac{2}{\left(n-1\right)n\left(n+1\right)}\)$
$\(\Rightarrow2A< \dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+……………….+\dfrac{2}{\left(n-1\right)n\left(n+1\right)}\)$
$\(2A< \dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+…………..+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\)$
$\(2A< \dfrac{1}{1.2}-\dfrac{1}{n\left(n+1\right)}\)$
$\(\Rightarrow A< \left(\dfrac{1}{1.2}-\dfrac{1}{n\left(n+1\right)}\right):2\)$
$\(A< \dfrac{1}{4}-\dfrac{1}{2n\left(n+1\right)}\)$
$\(\Rightarrow A< \dfrac{1}{4}\) \(\rightarrowđpcm\)$