부분분수 급수

공차가 $3$인 두 등차수열 $\left\{ a _{n} \right\}$, $\left\{ b _{n} \right\}$의 첫째항이 각각 $4$, $7$이므로 두 수열의 일반항은 $a _{n} = 4 + \left( n - 1 \right) \times 3 = 3 n + 1$ $b _{n} = 7 + \left( n - 1 \right) \times 3 = 3 n + 4$ 이때 $\displaystyle \frac{1}{a _{n} b _{n}} = \frac{1}{\left( 3 n + 1 \right) \left( 3 n + 4 \right)} = \frac{1}{3} \left( \frac{1}{3 n + 1} - \frac{1}{3 n + 4} \right)$ 이므로 급수 $\displaystyle \sum\limits _{n = 1} ^{\infty} \frac{1}{a _{n} b _{n}}$의 부분합을 $S _{n}$이라 하면 $S _{n}$$\displaystyle = \sum\limits _{k = 1} ^{n} \frac{1}{a _{k} b _{k}}$ $\displaystyle = \sum\limits _{k = 1} ^{n} \frac{1}{3} \left( \frac{1}{3 k + 1} - \frac{1}{3 k + 4} \right)$ $\displaystyle = \frac{1}{3} \left\{ \left( \frac{1}{4} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{10} \right) + \cdots \right.$$\displaystyle + \left( \frac{1}{3 n + 1} - \frac{1}{3 n + 4} \right) \}$ $\displaystyle = \frac{1}{3} \left( \frac{1}{4} - \frac{1}{3 n + 4} \right)$ $\therefore$ $\displaystyle \sum\limits _{n = 1} ^{\infty} \frac{1}{a _{n} b _{n}}$$\displaystyle = \lim\limits _{n \rightarrow \infty} S _{n}$ $\displaystyle = \frac{1}{3} \lim\limits _{n \rightarrow \infty} \left( \frac{1}{4} - \frac{1}{3 n + 4} \right)$ $\displaystyle = \frac{1}{3} \times \left( \frac{1}{4} - 0 \right)$ $\displaystyle = \frac{1}{12}$