Question

cho a = 1/2^3 + 1/3^3 + 1/4^3 + ...+ 1/2017^3 + 1/2018^3 . so sánh a với 1/4

Answer 1

Thừa số tổng quát: Với \(n\in N\circledast\)

\(\dfrac{1}{n^3}< \dfrac{1}{n^3-n}=\dfrac{1}{n\left(n^2-1\right)}=\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)

Thay vào bài toán:

\(a< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{2016.2017.2018}+\dfrac{1}{2017.2018.2019}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{2016.2017}-\dfrac{1}{2017.2018}+\dfrac{1}{2017.2018}-\dfrac{1}{2018.2019}\right)\)

\(=\dfrac{1}{4}-\dfrac{1}{2.2018.2019}< \dfrac{1}{4}\Leftrightarrow a< \dfrac{1}{4}\)