Question

chứng minh rằng:\(\frac{1}{2}^2+\frac{1}{3}^2+....+\frac{1}{100}^2< 1\)

Answer 1

Đặt :

\(A=\frac{1}{2^2}+\frac{1}{3^2}+.........+\frac{1}{100^2}\)

Ta thấy :

\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

................

\(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Leftrightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...........+\frac{1}{99.100}\)

\(\Leftrightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.......+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow A< 1-\frac{1}{100}< 1\)

\(\Leftrightarrow A< 1\)