Có $n^{2}>(n-1)(n+1)\Rightarrow \frac{1}{n^{2}}<\frac{1}{(n-1)(n+1)}$(1)Áp dụng (1) ta được: $VT<\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{(2n-1)(2n+1)}$
$\Leftrightarrow VT<\frac{1}{2}(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1})=\frac{1}{2}(\frac{1}{3}-\frac{1}{2n+1})<\frac{1}{2}.\frac{1}{3}<\frac{1}{4}$