Dễ chứng minh bằng quy nạp đẳng thức sau:
$ \frac{1^{2}}{1.3}+\frac{2^{2}}{3.5}+\frac{3^{2}}{5.7}+...+\frac{n^{2}}{(2n-1)(2n+1)} =\frac{n(n+1)}{4n+2}$
Suy ra
$\mathop {\lim }\left[ {\left ( \frac{1^{2}}{1.3}+\frac{2^{2}}{3.5}+\frac{3^{2}}{5.7}+...+\frac{n^{2}}{(2n-1)(2n+1)} \right )}\frac{1}{2} \right]=\lim \frac{n(n+1)}{8n+4}=\lim \frac{n+1}{8+\frac4n}=+\infty.$