证明：
$\begin{aligned}&\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+\cdots+\frac{1}{(2n - 1)\times(2n+1)}\\=&\frac{1}{2}\times\left(1-\frac{1}{3}\right)+\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}\times\left(\frac{1}{5}-\frac{1}{7}\right)+\cdots+\frac{1}{2}\times\left(\frac{1}{2n - 1}-\frac{1}{2n+1}\right)\\=&\frac{1}{2}\times\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\cdots+\left(\frac{1}{2n - 1}-\frac{1}{2n+1}\right)\right]\\=&\frac{1}{2}\times\left(1-\frac{1}{2n + 1}\right)\\=&\frac{1}{2}\times\frac{2n+1 - 1}{2n+1}\\=&\frac{n}{2n+1}\\\end{aligned}$
因为$n$为正整数，所以$\frac{n}{2n + 1}＜\frac{1}{2}。$
