Question

CMR: \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{99^2}\) < 2

 

Answer 1

\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{99^2}\)

\(A< \frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{98.99}\)

\(A< 1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{99}\)

\(A< 2-\frac{1}{99}< 2\)

Vậy A < 2 

Answer 2

\(\Rightarrow A< \frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

\(\Rightarrow A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}+\frac{1}{99}\)

\(\Rightarrow A< 2-\frac{1}{99}< 2\)

\(\Rightarrow A< 2\)