Question

Chứng minh rằng: C=4/5^2+4/6^2+......+4/2021^2<1/2

Answer 1

Sửa đề: Chứng minh C<1

Ta có: \(\frac{1}{5^2}<\frac{1}{4\cdot5}=\frac14-\frac15\)

\(\frac{1}{6^2}<\frac{1}{5\cdot6}=\frac15-\frac16\)

...

\(\frac{1}{2021^2}<\frac{1}{2020\cdot2021}=\frac{1}{2020}-\frac{1}{2021}\)

Do đó: \(\frac{1}{5^2}+\frac{1}{6^2}+\cdots+\frac{1}{2021^2}<\frac14-\frac15+\frac15-\frac16+\cdots+\frac{1}{2020}-\frac{1}{2021}=\frac14-\frac{1}{2021}<\frac14\)

=>\(4\left(\frac{1}{5^2}+\frac{1}{6^2}+\cdots+\frac{1}{2021^2}\right)<4\cdot\frac14\)

=>C<1