부분분수 무한급수

$\displaystyle \sum\limits _{k = 1} ^{n} \frac{1}{k ( k + 2 )} = \sum\limits _{k = 1} ^{n} \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k + 2} \right)$ $\displaystyle = \frac{1}{2} \left( \frac{1}{1} + \frac{1}{2} - \frac{1}{n + 1} - \frac{1}{n + 2} \right)$ $\displaystyle \sum\limits _{n = 1} ^{\infty} \frac{8}{n ( n + 2 )} = 8 \lim\limits _{n \rightarrow \infty} {} \sum\limits _{k = 1} ^{n} \frac{1}{k ( k + 2 )}$ $\displaystyle = 8 \lim\limits _{n \rightarrow \infty} {} \frac{1}{2} \left( \frac{1}{1} + \frac{1}{2} - \frac{1}{n + 1} - \frac{1}{n + 2} \right) = 6$