Simplify this summation

Question

I have $\frac{(m-1)!}{(m+n)!} \sum_{i=0}^{m-1} \frac{(m+n-i-1)!}{(m-1-i)!}$. And i'm fairly certain that this equals 1/(n+1) but i'm not sure how I would simplify the big summation term to show that.

Answer 1

It helps to write your sum as $\frac{(m-1)!n!}{(m+n)!}\sum_{j=0}^{m-1}\binom{n+j}{j}$, where we take $j=m-1-i$. With the famous result$$\sum_{j=0}^{m-1}\binom{n+j}{j}=\binom{m+n}{m-1},$$we get the desired$$\frac{(m-1)!n!(m+n)!}{(m+n)!(m-1)!(n+1)!}=\frac{n!}{(n+1)!}=\frac{1}{n+1}.$$