Problem - 4321
Prove $$\sum_{k=0}^{n}(-1)^k\frac{{n \choose k}}{\binom{m+k}{k}}=\frac{m}{m+n}$$
The conclusion of # 4303 states $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{m}{m+k}=\frac{1}{\binom{m+n}{n}}$$
Letting $a_n=\frac{m}{m+n}$ and $b_n=\binom{m+n}{n}^{-1}$ will yield $$b_n=\sum_{k=0}^{n}(-1)^k\binom{n}{k}a_n$$
Employing the inverse method will lead to desired result immediately.