a, Do \(\dfrac{1}{n^2}< \dfrac{1}{n^2-1}\) với mọi n ≥ 2 nên
A < C = \(\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+\dfrac{1}{4^2-1}+......+\dfrac{1}{n^2-1}\)
Mặc khác :
\(C=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+......+\dfrac{1}{\left(n-1\right).\left(n+1\right)}\)
\(C=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+......+\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)
\(C=-\left(1+\dfrac{1}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)< \dfrac{1}{2}.\dfrac{3}{2}=\dfrac{3}{4}< 1\)
Vậy A < 1
b,\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+.....+\dfrac{1}{\left(2n\right)^2}\)
\(B=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{n^2}\right)\)
\(B=\dfrac{1}{2^2}\left(1+A\right)\)
Suy ra \(P< \dfrac{1}{2^2}\left(1+1\right)=\dfrac{1}{2}\) ; hay \(P< \dfrac{1}{2}\)
