자연계열 1번

1. $\displaystyle k$번째 항은,$\displaystyle \frac{2k+1}{k(k+1)(k+2)}=\frac Ak+\frac B{k+1}+\frac C{k+2}$이므로, $\displaystyle A=\frac12,\ B=1,\ C=-\frac32$이다.

2. $\displaystyle \begin{aligned}S_n&=\left[\frac{\frac12}1+\frac12-\frac{\frac32}3\right]+\left[\frac{\frac12}2+\frac13-\frac{\frac32}4\right]+\left[\frac{\frac12}3+\frac14-\frac{\frac32}5\right]+\cdots\\&\quad+\left[\frac{\frac12}{n-2}+\frac1{n-1}-\frac{\frac32}n\right]+\left[\frac{\frac12}{n-1}+\frac1n-\frac{\frac32}{n+1}\right]+\left[\frac{\frac12}n+\frac1{n+1}-\frac{\frac32}{n+2}\right]\end{aligned}$부분 항들 중, $\displaystyle -\frac{\frac32}k+\frac1k+\frac{\frac12}k=0\ (k\ge3)$을 만족하는 세 짝을 찾을 수 있으므로 이것들을 상쇄하면,$\displaystyle \begin{aligned}S_n&=\left[\frac{\frac12}1+\frac12\right]+\left[\frac{\frac12}2\right]+\left[-\frac{\frac32}{n+1}\right]+\left[\frac1{n+1}-\frac{\frac32}{n+2}\right]=\frac54-\frac{\frac12}{n+1}-\frac{\frac32}{n+2}\\&=\frac54-\frac1{n^2+3n+2}\left(2n+\frac52\right)\end{aligned}$

3.$\displaystyle S=\frac12\left[\left(1-\frac13\right)\left(1+\frac12+\frac13\right)+\left(\frac12-\frac14\right)\left(1+\frac12+\frac13+\frac14\right)+\left(\frac13-\frac15\right)\left(1+\frac12+\frac13+\frac14+\frac15\right)+\cdots\right]$$\displaystyle \begin{aligned}2S&=\left(1+\frac12+\frac13\right)+\frac12\left(1+\frac12+\frac13+\frac14\right)\\&\quad-\frac13\left(1+\frac12+\frac13\right)+\frac13\left(1+\frac12+\frac13+\frac14+\frac15\right)\\&\quad-\frac14\left(1+\frac12+\frac13+\frac14\right)+\frac14\left(1+\frac12+\frac13+\frac14+\frac15+\frac16\right)\\&\quad\cdots\end{aligned}$$\displaystyle 2S=\left(1+\frac12+\frac13\right)+\frac12\left(1+\frac12+\frac13+\frac14\right)+\frac13\left(\frac14+\frac15\right)+\frac14\left(\frac15+\frac16\right)+\frac15\left(\frac16+\frac17\right)+\cdots$$\displaystyle \begin{aligned}E&=\frac13\left(\frac14+\frac15\right)+\frac14\left(\frac15+\frac16\right)+\frac15\left(\frac16+\frac17\right)+\cdots\\&=\left[\frac1{3\cdot4}+\frac1{4\cdot5}+\frac1{5\cdot6}+\cdots\right]+\left[\frac1{3\cdot5}+\frac1{4\cdot6}+\frac1{5\cdot7}+\cdots\right]\\&=\frac13+\frac12\left[\frac13+\frac14\right]=\frac13+\frac7{24}=\frac{15}{24}=\frac58\end{aligned}$