Question

chung minh1/2^2+1/4^2+1/6^2+...........+1/2016^2<1/2

Answer 1

\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{2016^2}\)

\(=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{1008^2}\right)< \frac{1}{2^2}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1007.1008}\right)\)

                                                                         \(< \frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1007}-\frac{1}{1008}\right)\)

                                                                           \(< \frac{1}{4}.\left(2-\frac{1}{1008}\right)< \frac{1}{4}.2=\frac{1}{2}\)

=> đpcm