ta thấy \(\frac{1}{2^2}=\frac{1}{2.2}<\frac{1}{1.2};\frac{1}{3^2}=\frac{1}{3.3}<\frac{1}{2.3};...;\frac{1}{n^2}=\frac{1}{n.n}<\frac{1}{\left(n-1\right).n}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}<\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{\left(n-1\right).n}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+..+\frac{1}{n^2}<1-\frac{1}{n}<1\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+..+\frac{1}{n^2}<1\left(đpcm\right)\)
tớ biết làm : 1 +1+....+1<1 +1 +....+1 =1-1 <1
22 32 n2 1.2 2.3 (n-1)n n