Question

          

Answer 1

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

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

\(B=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}=\frac{1}{\left(2.1\right)^2}+\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+...+\frac{1}{\left(2.n\right)^2}\)

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

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