Giải:

\(P=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}\left(n\in N,n\ge2\right).\)

\(P< \dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{\left(2n-2\right)2n}.\)

\(P< \dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right).\)

\(P< \dfrac{1}{2}\left[\dfrac{1}{2}+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+\left(\dfrac{1}{6}-\dfrac{1}{6}\right)+...+\left(\dfrac{1}{2n-2}-\dfrac{1}{2n-2}\right)-\dfrac{1}{2n}\right].\)

\(P< \dfrac{1}{2}\left[\dfrac{1}{2}+0+0+...+0-\dfrac{1}{2n}\right].\)

\(P< \dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2n}\right).\)

\(P< \dfrac{1}{2}.\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{2n}.\)

\(P< \dfrac{1}{4}-\dfrac{1}{2}.\dfrac{1}{2n}.\)

\(P< \dfrac{1}{4}-\dfrac{1}{\left(2n\right)2}< \dfrac{1}{4}.\)

Vậy ta thu được \(đpcm.\)

~ Học tốt!!! ~