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? 
 A: 
We obtain
  \begin{align*}
\color{blue}{\sum_{B=0}^{n-1}}&\color{blue}{\sum_{A=0}^{n-B-1}\binom{n-1}{B}\binom{n-1-B}{A}\frac{A}{A+B+1}}\\
&=\sum_{B=0}^{n-1}\binom{n-1}{B}\sum_{A=1}^{n-1-B}\binom{n-1-B}{A}A\int_{0}^{1}z^{A+B}\,dz\tag{1}\\
&=\int_{0}^{1}\sum_{B=0}^{n-1}\binom{n-1}{B}\sum_{A=1}^{n-1-B}\binom{n-2-B}{A-1}(n-1-B)z^{A+B}\,dz\tag{2}\\
&=\int_{0}^{1}\sum_{B=0}^{n-1}\binom{n-1}{B}(n-1-B)\sum_{A=0}^{n-2-B}\binom{n-2-B}{A}z^{A+B+1}\,dz\tag{3}\\
&=\int_{0}^{1}\sum_{B=0}^{n-1}\binom{n-1}{B}(n-1-B)z^{B+1}(1+z)^{n-2-B}\,dz\tag{4}\\  
&=\int_{0}^{1}\sum_{B=0}^{n-1}\binom{n-2}{B}(n-1)z^{B+1}(1+z)^{n-2-B}\,dz\tag{5}\\    
&=(n-1)\int_{0}^{1}z\sum_{B=0}^{n-2}\binom{n-2}{B}z^B(1+z)^{n-2-B}\,dz\\      
&=(n-1)\int_{0}^{1}z(1+2z)^{n-2}\,dz\tag{6}\\      
&=\frac{n-1}{4}\int_{1}^{3}(u-1)u^{n-2}\,du\tag{7}\\      
&=\frac{n-1}{4}\int_{1}^{3}\left(u^{n-1}-u^{n-2}\right)\,du\\      
&=\frac{n-1}{4}\left.\left(\frac{1}{n}u^n-\frac{1}{n-1}u^{n-1}\right)\right|_{1}^{3}\\
&=\frac{n-1}{4}\left(\frac{3^n-1}{n}-\frac{3^{n-1}-1}{n-1}\right)\\
&\,\,\color{blue}{=\frac{3^{n-1}}{2}-\frac{3^n-1}{4n}}
\end{align*}

Comment:


*

*In (1) we use the integral to get rid of the denominator. We also start the inner sum with $A=1$.

*In (2) we use the binomial identity $\binom{p}{q}=\binom{p-1}{q-1}\frac{p}{q}$.

*In (3) we shift the index of the inner sum to start with $A=0$.

*In (4) we apply the binomial theorem.

*In (5) we use the binomial identity $\binom{p}{q}=\binom{p-1}{p-q-1}\frac{p}{p-q}$.

*In (6) we use again the binomial theorem.

*In (7) we substitute $1+2z=u, 2dz=du$.
A: Wolfram Alpha found this sequence which seems to match the sum for all positive integers...
$$ a_n = \frac{3^{n-1}}{2} - \frac{3^n - 1}{4n} $$
