Giving a closed expression to $\sum_{i=0}^b (-1)^{b-i} \binom{b}{i}\frac{1}{a+b-i}$

Question

I want to prove $\sum_{i=0}^b (-1)^{b-i} \binom{b}{i}\frac{1}{a+b-i} = \frac{(a-1)! b!}{(a+b)!}$ yet I feel like I don't know how to even approach this problem.

Any hints are welcome.

Answer 1

Note that $$\eqalign{\sum_{i=0}^b(-1)^{b-i}\binom{b}{i}\frac{1}{a+b-i}&= \int_0^1\sum_{i=0}^b(-1)^{b-i}\binom{b}{i}x^{a+b-i-1}dx\cr &=\int_0^1x^{a-1}(1-x)^b dx\cr &=\frac{\Gamma(a)\Gamma(b+1)}{\Gamma(a+b+1)}=\frac{(a-1)! b!}{(a+b)!} }$$