Question

[tran minh anh]

Với mọi số tự nhiên n \(\ge2\)hãy so sánh

A=1/2² + 1/3² + 1/4² + .....+ 1/n²  với 1

B=1/2² + 1/4² + 1/6² + .....+ 1/(2n)²  với 1/2

Answer 1

[Huỳnh Quang Sang]

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)

\(A=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{n\cdot n}\)

\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{(n-1)\cdot n}\)

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

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

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

Answer 2

[Huỳnh Quang Sang]

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

Theo câu a \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\) nên \(B< \frac{1}{4}\cdot1=\frac{1}{4}\)