Question

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

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

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

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

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

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

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

Answer 2

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

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

Answer 3

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

Vậy A<1

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

Vậy B<\(\frac{1}{2}\)

   

Answer 4

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

Vậy...

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