Question

\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)

Chứng minh A<2

Answer 1

Ta có 

\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3}......\frac{1}{50^2}<\frac{1}{49.50}\)

\(=>A<1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)

=> A<2-1/50

=> A < 2

=> đpcm

Answer 2

Ta có: A = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)

 A < \(\frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\) 

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

A< 1 +1 -\(\frac{1}{50}\)

A< 2 - \(\frac{1}{50}\)

Vậy A< 2