ta có :
\(P=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}=\frac{1}{4}\times\left(\frac{1}{1^2}+\frac{1}{2^2}+..+\frac{1}{50^2}\right)\)
\(< \frac{1}{4}\times\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)=\frac{1}{4}\times\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=\frac{1}{4}\times\left(2-\frac{1}{50}\right)< \frac{1}{4}\times2=\frac{1}{2}\)
Vậy \(P< \frac{1}{2}\)