Question

Với mọi số tự nhiên n≥2, hãy so sánh:

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

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

Answer 1

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)

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

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

Vậy \(A< 1\)

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

Mà \(A< 1\Rightarrow B< \frac{1}{2^2}\left(1+1\right)=\frac{1}{2}\)

\(\Rightarrow B< \frac{1}{2}\)