자연계열 2번

합의 순서를 바꾸면$\displaystyle \begin{aligned}\sum_{n=1}^{10}a_n&=\sum_{m=1}^{10}\frac{\sum_{n=1}^m(m+2n-1)}{m^3(m+1)(m+2)}\\&=\sum_{m=1}^{10}\frac{2m^2}{m^3(m+1)(m+2)}\\&=\sum_{m=1}^{10}\left\{\frac1{m(m+1)}-\frac1{(m+1)(m+2)}\right\}\\&=\frac12-\frac1{132}=\frac{65}{132}\end{aligned}$이다.