급수의 수렴과 합

주어진 급수 $\displaystyle \sum\limits _{n = 1} ^{\infty} \left( \frac{a - 3 n}{n} + \frac{an + 6}{n + a} \right)$이 수렴하려면 $\displaystyle \lim\limits _{n \rightarrow \infty} {} \left( \frac{a - 3 n}{n} + \frac{an + 6}{n + a} \right) = 0$ 이어야 한다. 이때 $\displaystyle \lim\limits _{n \rightarrow \infty} {} \left( \frac{a - 3 n}{n} + \frac{an + 6}{n + a} \right)$$\displaystyle = \lim\limits _{n \rightarrow \infty} {} \left( \frac{\frac{a}{n} - 3}{1} + \frac{a + \frac{6}{n}}{1 + \frac{a}{n}} \right)$ $= - 3 + a$ 이므로 $- 3 + a = 0$ $\therefore$ $a = 3$ 즉 $\displaystyle \sum\limits _{n = 1} ^{\infty} \left( \frac{3 - 3 n}{n} + \frac{3 n + 6}{n + 3} \right)$ $\displaystyle = \sum\limits _{n = 1} ^{\infty} \left\{ \frac{3}{n} - 3 + \frac{3 \left( n + 3 \right) - 3}{n + 3} \right\}$ $\displaystyle = \sum\limits _{n = 1} ^{\infty} \left( \frac{3}{n} - \frac{3}{n + 3} \right)$ $\displaystyle = 3 \lim\limits _{n \rightarrow \infty} {} \sum\limits _{k = 1} ^{n} \left( \frac{1}{k} - \frac{1}{k + 3} \right)$ $\displaystyle = 3 \lim\limits _{n \rightarrow \infty} {} \left\{ \left( 1 - \frac{1}{4} \right) + \left( \frac{1}{2} - \frac{1}{5} \right) + \left( \frac{1}{3} - \frac{1}{6} \right) \right.$$\displaystyle + \left( \frac{1}{4} - \frac{1}{7} \right) + \left( \frac{1}{5} - \frac{1}{8} \right) + \cdots$ $\displaystyle + \left( \frac{1}{n - 2} - \frac{1}{n + 1} \right) + \left( \frac{1}{n - 1} - \frac{1}{n + 2} \right)$$\displaystyle \left. + \left( \frac{1}{n} - \frac{1}{n + 3} \right) \right\}$ $\displaystyle = 3 \lim\limits _{n \rightarrow \infty} {} \left( 1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{n + 1} - \frac{1}{n + 2} - \frac{1}{n + 3} \right)$ $\displaystyle = 3 \times \frac{6 + 3 + 2}{6}$ $\displaystyle = \frac{11}{2}$ 따라서 $\displaystyle S = \frac{11}{2}$이므로 $\displaystyle a + S = 3 + \frac{11}{2} = \frac{17}{2}$