\(\frac{3}{1^2x2^2}\)+\(\frac{5}{2^2x3^2}\)+...+\(\frac{39}{19^2x20^2}\)<1
=\(\frac{3}{1.4}\)+\(\frac{5}{4x9}\)+...+\(\frac{39}{361x400}\)<1
=1-\(\frac{1}{4}\)+\(\frac{1}{4}\)-...-\(\frac{1}{361}\)+\(\frac{1}{361}\)-\(\frac{1}{400}\)<1
vì 1-\(\frac{1}{400}\)<1 nên \(\frac{3}{1^2x2^2}\)+\(\frac{5}{2^2x3^2}\)+...+\(\frac{39}{39^2x40^2}\)<1
vậy..............................................