回復 1# 阿良 的帖子
\(\displaystyle a=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{1}{2003}-\frac{1}{2004}\)
\(\displaystyle =\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{2004}-2\Big(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{2004}\Big)\)
\(\displaystyle =\frac{1}{1003}+\frac{1}{1004}+\cdots +\frac{1}{2004}=\frac{3007}{1003\times2004}+\frac{3007}{1004\times2003}+\cdots +\frac{3007}{1503\times1504}=3007\Big(\frac{b}{2}\Big)\)
\(\displaystyle \Rightarrow \frac{a}{b}=\frac{3007}{2}\)
