부분분수 급수

$\displaystyle \frac{2 n + 3}{n ( n + 1 ) ( n + 2 )}$$\displaystyle = \frac{( n + 2 )}{n ( n + 1 ) ( n + 2 )} + \frac{( n + 1 )}{n ( n + 1 ) ( n + 2 )}$ $\displaystyle = \frac{1}{n ( n + 1 )} + \frac{1}{n ( n + 2 )}$$\displaystyle = \left( \frac{1}{n} - \frac{1}{n + 1} \right) + \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n + 2} \right)$ $\therefore$$\displaystyle \sum\limits _{n = 1} ^{\infty} \frac{2 n + 3}{n ( n + 1 ) ( n + 2 )}$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {} \sum\limits _{k = 1} ^{n} \left\{ \left( \frac{1}{k} - \frac{1}{k + 1} \right) + \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k + 2} \right) \right\}$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {\left\{ \left( 1 - \frac{1}{n + 1} \right) + \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{n + 1} - \frac{1}{n + 2} \right) \right\}}$ $\displaystyle = 1 + \frac{3}{4}$$\displaystyle = \frac{7}{4}$ $\therefore$$p + q = 11$