TITLE: Orthogonal combinatoric sum
QUESTION [6 upvotes]: I have verified this identity in Matlab:
$$ \sum_{k=m}^n~(-)^{n+k}\frac{2k+1}{n+k+1}\binom{n}{k}\binom{n+k}{k}^{-1}\binom{k}{m}\binom{k+m}{m}=\delta_{nm} $$
Where $n, m$ are positive integers, and where $(-)^p$ means $(-1)^p$.
It was found by combining two inverse expansions $$ f_n=\sum_{k=0}^n\frac{2k+1}{n+k+1}\binom{n}{k}\binom{n+k}{k}^{-1}g_k  $$ and
$$ g_n =\sum_{k=0}^n(-)^{n-k}\binom{n}{k}\binom{n+k}{k}f_k $$
where $f_n$ and $g_n$ are orthogonal polynomials (related to Legendre polynomials) of several variables.
Could anyone offer a proof of this? Or recommend any books etc. that might be useful?

REPLY [2 votes]: Suppose we seek to verify that
$$\sum_{k=m}^n (-1)^{n+k} \frac{2k+1}{n+k+1}
{n\choose k} {n+k\choose k}^{-1} {k\choose m} {k+m\choose m}
= \delta_{mn}.$$
Here we may assume $n\ge m$, the equality holds trivially otherwise.
Now we have
$${n\choose k}{n+k\choose k}^{-1}
= \frac{n!}{k! (n-k)} \frac{k! n!}{(n+k)!}
\\ = \frac{n!}{(n-k)} \frac{n!}{(n+k)!}
= {2n\choose n+k} {2n\choose n}^{-1}.$$
We get for the sum
$$\sum_{k=m}^n (-1)^{n+k} \frac{2k+1}{n+k+1}
{2n\choose n+k} {k\choose m} {k+m\choose m}
= \delta_{mn} \times {2n\choose n}.$$
which is
$$\sum_{k=m}^n (-1)^{n+k} (2k+1)
{2n+1\choose n+k+1} {k\choose m} {k+m\choose m}
\\ = \delta_{mn} \times (2n+1)\times {2n\choose n}.$$
Introduce
$${2n+1\choose n+k+1} = {2n+1\choose n-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} (1+z)^{2n+1} \; dz.$$
Observe that this  vanishes when $k\gt n$ so we  may extend $k$ upward
to infinity.

Furthermore introduce
$${k\choose m}
= \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{m+1}} (1+w)^k \; dw.$$
Observe once  again that the  integral vanishes, this time  when $0\le
k\lt m$ so we may extend $k$ back to zero.

We thus get for the sum
$$(-1)^n \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{2n+1} 
\\ \times \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{m+1}} 
\sum_{k\ge 0}  (-1)^k (2k+1) {k+m\choose m} z^k (1+w)^k
\; dw\; dz.$$
The inner sum yields two pieces, the first is
$$\sum_{k\ge 0}  (-1)^k {k+m\choose m} z^k (1+w)^k
= \frac{1}{(1+z+wz)^{m+1}}
\\ = \frac{1}{(1+z)^{m+1}} \frac{1}{(1+wz/(1+z))^{m+1}}.$$
On extracting the residue for the integral in $w$ we obtain
$$(-1)^n \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{2n+1} 
\\ \times 
\frac{1}{(1+z)^{m+1}} {2m\choose m} (-1)^m 
\frac{z^m}{(1+z)^m} \; dz
\\ = {2m\choose m} (-1)^{n+m}
\int_{|z|=\epsilon} \frac{1}{z^{n-m+1}}
(1+z)^{2n-2m} \; dz
\\ = {2m\choose m} (-1)^{n+m} {2n-2m\choose n-m}.$$
The second piece from the sum is
$$2 \sum_{k\ge 1}  (-1)^k k {k+m\choose m} z^k (1+w)^k.$$
Write $$k{k+m\choose m} = \frac{(k+m)!}{(k-1)! m!}
= (m+1) \frac{(k+m)!}{(k-1)! (m+1) !}
\\ = (m+1){k+m\choose m+1}$$
to get for the sum
$$2(m+1) z(1+w) 
\sum_{k\ge 1}  (-1)^k {k+m\choose m+1} z^{k-1} (1+w)^{k-1}
\\ = - 2(m+1)z(1+w) \frac{1}{(1+z+wz)^{m+2}}
\\ = - 2(m+1)z(1+w) 
\frac{1}{(1+z)^{m+2}} \frac{1}{(1+wz/(1+z))^{m+2}}.$$
Here we get two pieces, the first is
$$- 2(m+1) (-1)^n \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{z}{z^{n+1}} (1+z)^{2n+1} 
\\ \times 
\frac{1}{(1+z)^{m+2}} {2m+1\choose m} (-1)^m 
\frac{z^m}{(1+z)^m} \; dz
\\ = -2(m+1) {2m+1\choose m} (-1)^{n+m}
\int_{|z|=\epsilon} \frac{1}{z^{n-m}}
(1+z)^{2n-2m-1} \; dz$$
We have two cases, we get zero when $n=m$
and when $n\gt m$ we have
$$ -2(m+1){2m+1\choose m} (-1)^{n+m} {2n-2m-1\choose n-m-1}.$$
The second piece is
$$- 2(m+1) (-1)^n \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{z}{z^{n+1}} (1+z)^{2n+1} 
\\ \times 
\frac{1}{(1+z)^{m+2}} {2m\choose m-1} (-1)^{m-1}
\frac{z^{m-1}}{(1+z)^{m-1}} \; dz
\\ = 2(m+1) {2m\choose m-1} (-1)^{n+m}
\int_{|z|=\epsilon} \frac{1}{z^{n-m+1}}
(1+z)^{2n-2m} \; dz
\\ = 2(m+1) {2m\choose m-1} (-1)^{n+m} {2n-2m\choose n-m}.$$
Therefore when $n=m$ we get
$${2n-2m\choose n-m} (-1)^{m+n}
\left(2(m+1){2m\choose m-1} + {2m\choose m}\right).$$
This simplifies to
$$(-1)^{2m} 
\left(2(m+1){2m\choose m-1} + {2m\choose m}\right)
\\ = 2m{2m\choose m} + {2m\choose m}
= (2m+1){2m\choose m}.$$
This is precisely the claim we were trying to prove. On the other hand
when $n\gt m$ we obtain
$${2n-2m\choose n-m} (-1)^{m+n} \\ \times
\left(2(m+1){2m\choose m-1} + {2m\choose m}
- 2(m+1){2m+1\choose m} \frac{n-m}{2n-2m}\right).$$
The factor is
$$(2m+1){2m\choose m} - (m+1){2m+1\choose m} = 0.$$
This concludes the argument.
Remark. For  $n=m$ we could  have evaluated the single  term in
the  initial  sum by  expanding  the  four  binomial coefficients  and
assumed $n\gt m$ thereafter.