Question

Cho B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{100^2}\). So sánh B với 3/4

Answer 1

Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{4}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}\)

.......

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow B< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{3}{4}-\dfrac{1}{100}< \dfrac{3}{4}\)

Vậy B<3/4