\(B=\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}\)
\(=1+\frac{98}{2}+1+\frac{97}{3}+...+1+\frac{1}{99}+1\)
\(=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}\)
\(=100.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=100A\)
\(\Rightarrow\frac{A}{B}=\frac{A}{100A}=\frac{1}{100}\)