Double sum factorial manipulation $$\sum_{B = 0}^{n-1} \sum_{A = 0}^{n-B-1} \frac{(n-1)!}{B!(n-B-1)!} \frac{(n-B-1)!}{A!(n-B-A-1)!} \frac{A}{A+B+1}$$
This is driving me nuts! Is there anyway to reduce
$$\sum_{B = 0}^{n-1} \sum_{A = 0}^{n-B-1} \frac{(n-1)!}{A!B!(n-A-B-1)!} \frac{A}{A+B+1}$$
beyond
$$\sum_{B = 0}^{n-1} \sum_{A = 0}^{n-B-1} \frac{(n-1)!}{(A-1)!B!(n-A-B-1)!} \frac{1}{A+B+1}$$
I can't figure out how to deal with that $\frac{1}{A+B+1}$ term in any way that brings it inside the $\frac{1}{n-A-B-1}$ term. Is this not possible? 

Wolfram Alpha found this sequence which seems to match the sum for all positive integers...
$$ a_n = \frac{3^{n-1}}{2} - \frac{3^n - 1}{4n} $$
