\(A=\frac{1}{4}+\frac{1}{9}+...+\frac{1}{81}+\frac{1}{100}\)
\(=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}+\frac{1}{10.10}>\frac{1}{2^2}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(>\frac{1}{4}+\frac{1}{3}-\frac{1}{11}=\frac{65}{132}\)
\(\Rightarrow\) \(A>\frac{65}{132}\)