Question

Với mọi số tự nhiên n >= 2 hãy so sánh:

a)A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{n^2}\) với 1

b)B=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+....+\dfrac{1}{\left(2n\right)^2}\) với \(\dfrac{1}{2}\)

Answer 1

a/ \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\)

Vậy A < 1

b/ Dựa vô câu a mà làm câu b nhé

\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{\left(2n\right)^2}=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)

\(< \dfrac{1}{4}\left(1+1-\dfrac{1}{n}\right)=\dfrac{1}{2}-\dfrac{1}{4n}< \dfrac{1}{2}\)

Vậy \(B< \dfrac{1}{2}\)