호 이등분점 부채꼴 극한

[출제의도] 규칙을 발견하여 무한급수의 합을 구할수 있는가를 묻는 문항이다. $\displaystyle S _{n} = \frac{1}{2} \cdot 2 ^{2} \cdot \mathrm{\angle} A O P _{\mathit{n}}$ 이므로 $\displaystyle \lim\limits _{n \rightarrow \infty} {} S _{n} = 2 \lim\limits _{n \rightarrow \infty} {} \mathrm{\angle} A O P _{\mathit{n}}$ 한편, $\displaystyle \mathrm{\angle} AOP _{1} = \frac{\pi}{2}$ , $\displaystyle \mathrm{\angle} AOP _{2} = \frac{\pi}{2} + \frac{\pi}{4}$ , $\displaystyle \mathrm{\angle} AOP _{3} = \frac{\pi}{2} + \frac{\pi}{4} - \frac{\pi}{8}$ , $\displaystyle \mathrm{\angle} AOP _{4} = \frac{\pi}{2} + \frac{\pi}{4} - \frac{\pi}{8} + \frac{\pi}{16}$ , ⋮ ∴$\displaystyle \mathrm{\lim\limits} _{n \rightarrow \infty} {} \angle AOP _{\mathit{n}} = \frac{\pi}{2} + \frac{\pi}{4} - \frac{\pi}{8} + \frac{\pi}{16} - \cdots = \frac{\pi}{2} + \frac{\frac{\pi}{4}}{1 - \left( - \frac{1}{2} \right)}$ $\displaystyle = \frac{\pi}{2} + \frac{\pi}{6}$ $\displaystyle = \frac{2 \pi}{3}$ 따라서, $\displaystyle \lim\limits _{n \rightarrow \infty} {} S _{n} = 2 \lim\limits _{n \rightarrow \infty} {} \mathrm{\angle} AOP _{\mathit{n}} = \frac{4 \pi}{3}$