Question

[Ngọc Nguyễn Minh]

Với mọi STN \(n\ge2\) hãy so sánh

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

Answer 1

[Hoàng Phúc]

a)\(\frac{1}{2^2}=\frac{1}{2.2}<\frac{1}{1.2};\frac{1}{3^2}=\frac{1}{3.3}<\frac{1}{2.3};...;\frac{1}{n^2}=\frac{1}{n.n}<\frac{1}{\left(n-1\right).n}\)

=>\(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{\left(n-1\right).n}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{n-1}-\frac{1}{n}\)

=>A<1-1/n

mà 1-1/n<1

=>A<1

b)tương tự

Answer 2

[Giáp Ánh]

a)Ta có: A= \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{n^2}\Rightarrow A<\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n-1\right)n}\)

\(\Rightarrow\)\(A<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)

\(\Rightarrow A<1-\frac{1}{n}\Rightarrow A<1\)