Question

cho A =\(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{2019^2}\). So sánh A với \(\frac{3}{4}\)

GIÚP NHA!!!!!!!!!!!!!!!!!!!!! YÊU NHÌU

Answer 1

\(A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}=\frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)

=> \(A< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{2019}=\frac{3}{4}-\frac{1}{2019}=\frac{3}{4}\)

Vậy A<3/4

Answer 2

A< \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2018.2019}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)

=\(1-\frac{1}{2019}=\frac{2019-1}{2019}=\frac{2018}{2019}\)