Question

cho

\(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}\)

chứng tỏ tổng đó bé hơn 2

Answer 1

Gọi \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)là A

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}\)

           \(\frac{1}{3^2}<\frac{1}{2\cdot3}\)

            ...................

            \(\frac{1}{50^2}<\frac{1}{49\cdot50}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}<\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)

Ta có: \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}<1\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}<1\)=> \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}<2\)