망원급수와 공차

[출제의도] 급수의 뜻 이해하기 등차수열 $\left\{ a _{n} \right\}$의 일반항은 $a _{n} = 1 + ( n - 1 ) d$ $\displaystyle \sum\limits _{n = 1} ^{\infty} \left( \frac{n}{a _{n}} - \frac{n + 1}{a _{n + 1}} \right)$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {} \sum\limits _{k = 1} ^{n} \left( \frac{k}{a _{k}} - \frac{k + 1}{a _{k + 1}} \right)$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {} \left\{ \left( \frac{1}{a _{1}} - \frac{2}{a _{2}} \right) + \left( \frac{2}{a _{2}} - \frac{3}{a _{3}} \right) + \cdots + \left( \frac{n}{a _{n}} - \frac{n + 1}{a _{n + 1}} \right) \right\}$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {} \left( 1 - \frac{n + 1}{a _{n + 1}} \right)$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {} \left( 1 - \frac{n + 1}{dn + 1} \right)$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {} \left( 1 - \frac{1 + \frac{1}{n}}{d + \frac{1}{n}} \right)$ $\displaystyle = 1 - \frac{1}{d} = \frac{2}{3}$ 따라서 $d = 3$