Cho \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}\)... là A, ta có:
A = \(\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
A = \(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{2^2}+...\frac{1}{9^2}-\frac{1}{10^2}\)
A = 1 \(-\frac{1}{10^2}\) <1
Vậy: A < 1
\(\frac{3}{1^2.2^2}\)+\(\frac{5}{2^2.3^2}\)+...+\(\frac{19}{9^2.10^2}\)
=1-1/4+1/4-1/9+...1/81-1/100
=1-1/100<1
Vậy tổng trên <1
\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{19}{9^2.10^2}\\ =\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{19}{81.100}\\ =1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+...+\left(\frac{1}{81}-\frac{1}{81}\right)-\frac{1}{100}\\ =1-\frac{1}{100}< 1\\ \Rightarrow A< 1\left(đpcm\right)\)
