Reducing the form of $2\sum\limits_{j=0}^{n-2}\sum\limits_{k=1}^n {{k+j}\choose{k}}{{2n-j-k-1}\choose{n-k+1}}$.

Question

I've been toying around with simplifying the expression $2\sum\limits_{j=0}^{n-2}\sum\limits_{k=1}^n {{k+j}\choose{k}}{{2n-j-k-1}\choose{n-k+1}}$ (for integer only $n$) for a while, as I was hoping it would have some sort of simpler closed form representation, and so I plugged it into Mathematica's FullSimplify function and got back the expression: $$-\dfrac{2\pi n \csc (2 \pi n)}{\Gamma (n-1) \Gamma (-2 n) \Gamma (n+3)}$$ Which can be rewritten as $$-\dfrac{4\pi n^2 \csc (2 \pi n)}{2n\Gamma (n-1) \Gamma (-2 n) \Gamma (n+3)}$$ and then simplified via the property that $\Gamma (x) \Gamma (-x) = -\dfrac{\pi \csc (\pi x)}{x}$, as $$-\dfrac{4\pi n^2 \csc (2 \pi n)}{2n\Gamma (n-1) \Gamma (-2 n) \Gamma (n+3)}$$ $$=-\dfrac{\pi \csc (\pi 2n)}{2n} \dfrac{4n^2}{\Gamma (n-1) \Gamma (-2 n) \Gamma (n+3)}$$ $$=\Gamma (2n) \Gamma (-2n) \dfrac{4n^2}{\Gamma (n-1) \Gamma (-2 n) \Gamma (n+3)}$$ $$= \dfrac{4n^2\Gamma (2n)}{\Gamma (n-1) \Gamma (n+3)}$$ and as $n$ is strictly an integer, the expression can be rewritten as $$\dfrac{4n^2(2n-1)!}{(n-2)!(n+2)!}$$ $$= \dfrac{2n(2n)!}{(n-2)!(n+2)!}$$ $$ = 2n {{2n}\choose{n-2}}$$ My problem here is that I have no idea how to get to that first step without the use of Mathematica. I wouldn't even know where to start with this thing if I didn't have the computer's help to break it down first for me. I would very much appreciate advice about how I could solve this by hand, or a hint about how to start reducing it from its double sum form, if that's possible. I would also appreciate feedback about the correctness of my simplification. Thank you in advance.

Answer 1

$\quad \ 2\sum\limits_{j=0}^{n-2}\sum\limits_{k=1}^n {{k+j}\choose{k}}{{2n-j-k-1}\choose{n-k+1}}\\ =2\sum\limits_{k=1}^n \sum\limits_{j=0}^{n-2} {{k+j}\choose{k}}{{2n-j-k-1}\choose{n-k+1}}\\ = 2\sum\limits_{k=1}^n \sum\limits_{z = k}^{n - 2 + k} {z \choose k} {2n - z - 1 \choose n -k + 1} \\ = 2\sum\limits_{k=1}^n {2n \choose n + 2} \\ = 2n {2n \choose n + 2} $

The first $=$ is obtained by exchanging the order of two $\sum$s, while the second is obtained by letting $z = j + k$.

The third can be easily proved by a combinatoric proof. Suppose there are $2n$ balls numbered from $1, 2, \cdots, 2n$. There are totally ${2n \choose n + 2}$ ways to choose $n + 2$ balls from the collection. But here let's count another way. There are totally ${z \choose k}{2n - z - 1 \choose n - k + 1}$ ways to select $n + 2$ balls from the collection such that the $k + 1$th smallest number among the chosen balls is $z + 1$. By varying $z$ and summing up the # of ways, the following equation holds: $$ \sum\limits_{z=1}^{2n} {z \choose k}{2n -z -1 \choose n - k + 1} = {2n \choose n + 2} $$ But since ${z \choose k}{2n - z - 1 \choose n - k + 1} = 0$ when $z < k$ or $z > n + k - 2$, the third $=$ is obtained.

Answer 2

Alternate solution.

As before we start trying to evaluate $$S(n) = \sum_{q=0}^{n-2} \sum_{k=1}^n {k+q\choose k} {2n-q-k-1\choose n-k+1}$$

which we re-write as $$-\sum_{q=0}^{n-2} {2n-q-1\choose n+1} -\sum_{q=0}^{n-2} {n+1+q\choose n+1} + \sum_{q=0}^{n-2} \sum_{k=0}^{n+1} {k+q\choose k} {2n-q-k-1\choose n-k+1}.$$

Call these pieces up to sign from left to right $S_1, S_2$ and $S_3.$ The two pieces in front cancel the quantities introduced by extending $k$ to include the values zero and $n+1.$

Evaluation of $S_1.$

Introduce $${2n-q-1\choose n+1} = {2n-q-1\choose n-q-2} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n-q-1}}{z^{n-q-1}} \; dz.$$

This vanishes when $q\gt n-2$ so we may extend the sum to infinity to get $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n-1}}{z^{n-1}} \sum_{q\ge 0} \frac{z^q}{(1+z)^q}\; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n-1}}{z^{n-1}} \frac{1}{1-z/(1+z)} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n-1}} \; dz \\ = {2n\choose n-2}.$$

Evaluation of $S_2.$

Introduce $${n+1+q\choose n+1} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n+1+q}}{z^{n+2}} \; dz.$$

This yields for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^{n+2}} \sum_{q=0}^{n-2} (1+z)^q \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^{n+2}} \frac{(1+z)^{n-1}-1}{1+z-1} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^{n+3}} ((1+z)^{n-1}-1) \; dz \\ = {2n\choose n+2}.$$

A more efficient evaluation is to notice that when we re-index $q$ as $n-2-q$ in $S_2$ we obtain

$$\sum_{q=0}^{n-2} {n+1+n-2-q\choose n+1} = \sum_{q=0}^{n-2} {2n-q-1\choose n+1}$$

which is $S_1.$

Evaluation of $S_3.$

Introduce $${2n-q-k-1\choose n-k+1} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n-q-k-1}}{z^{n-k+2}} \; dz.$$

This effectively controls the range so we can let $k$ go to infinity to get

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n-1}}{z^{n+2}} \sum_{q=0}^{n-2} \sum_{k\ge 0} {k+q\choose q} \frac{z^k}{(1+z)^{q+k}} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n-1}}{z^{n+2}} \sum_{q=0}^{n-2} \frac{1}{(1+z)^q} \frac{1}{(1-z/(1+z))^{q+1}} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+2}} \sum_{q=0}^{n-2} \frac{1}{(1+z)^{q+1}} \frac{1}{(1-z/(1+z))^{q+1}} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+2}} \times (n-1) \times \; dz \\ = (n-1) \times {2n\choose n+1}.$$

Finally collecting the three contributions we obtain $$(n-1) \times {2n\choose n+1} - 2{2n\choose n+2} = (n+2) {2n\choose n+2} - 2{2n\choose n+2} \\ = n\times {2n\choose n+2}.$$