Ta có: \(n^2>n^2-1=n^2-n+n-1=\left(n+1\right)\left(n-1\right)\)
Lúc đó:
\(B=\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2019^3}\)
\(< \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2018.2019.2020}\)
\(2B< \frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2018.2019.2020}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2018.2019}-\frac{1}{2019.2020}\)
\(=\frac{1}{2}-\frac{1}{2019.2020}< \frac{1}{2}\)
\(2B< \frac{1}{2}\Rightarrow B< \frac{1}{2^2}\)
Vậy \(B=\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2019^3}< \frac{1}{2^2}\left(đpcm\right)\)
thank you bn nha
