Đặt :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+.........+\frac{1}{100^2}\)
Ta thấy :
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
...................
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Leftrightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\)
\(\Leftrightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A=\left(\frac{1}{2^2}\right)\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Đặt \(C=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(.......\)
\(\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow C< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow C< 2-\frac{1}{50}=\frac{99}{50}\)
\(\Rightarrow A=\frac{1}{4}.\frac{99}{50}=\frac{99}{200}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
