$\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}=\sum_{n=8}^{11}\frac{1}{n(n+1)}$
$=\sum_{n=1}^{4}\frac{1}{(n+7)(n+8)}=\sum_{n=1}^{4}(\frac{1}{n+7}-\frac{1}{n+8})$
$=\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}$
$=\frac{1}{8}-\frac{1}{12}=\frac{1}{24}$ olacaktır.