Question

[titanic]

\(A=\frac{1}{\sqrt{1.2018}}+\frac{1}{\sqrt{2.2017}}+....+\frac{1}{\sqrt{k.\left(2018-k+1\right)}}+....+\frac{1}{\sqrt{2018.1}}\)

So sánh \(A\) với \(2.\frac{2018}{2019}\)

Answer 1

[alibaba nguyễn]

\(\frac{1}{\sqrt{k\left(2018-k+1\right)}}>\frac{2}{k+2019-k}=\frac{2}{2019}\)

Ap dụng bài toan được

\(A>\frac{2}{2019}+\frac{2}{2019}+...+\frac{2}{2019}=2.\frac{2018}{2019}\)