Ta có: \(\dfrac{2n+1}{n^2(n+1)^2}=\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}, \ \forall n \ne \{0;-1\}\)
Khi đó: \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2.10^2}=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(=\dfrac{1}{1^2}-\dfrac{1}{10^2}=1-0,01=0,99<1\)
Vậy mệnh đề đã được chứng minh.