Double sum factorial manipulation

Question

$$\sum_{B = 0}^{n-1} \sum_{A = 0}^{n-B-1} \frac{(n-1)!}{B!(n-B-1)!} \frac{(n-B-1)!}{A!(n-B-A-1)!} \frac{A}{A+B+1}$$

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?

Answer 1

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 $1+2z=u, 2dz=du$.

Answer 2

Wolfram Alpha found this sequence which seems to match the sum for all positive integers...

$$ a_n = \frac{3^{n-1}}{2} - \frac{3^n - 1}{4n} $$