Alternating sum over triple products of binomial coefficients divided by $2k+1$

Question

Let $T(n,m)$ be a rational coefficients such that $$ T(n,m) = \sum\limits_{k=0}^{m} \binom{n+m+1}{m+k+1} \binom{n+m-k}{n} \binom{m+k+1}{k} \frac{(-1)^k}{2k+1}. $$

I conjecture that for $n \geqslant 0, m \geqslant 0$ we have $$ T(n,m) = \frac{1}{2n+1} \binom{2(n+m+1)}{2(m+1)}. $$

Here is the PARI/GP program to check it numerically:

T(n,m) = {sum(k=0, m, 
binomial(n+m+1,m+k+1)
*binomial(n+m-k,n)
*binomial(m+k+1,k)
*(-1)^k/(2*k+1))}
test(n,m) = {T(n,m) == 
binomial(2*(n+m+1),2*(m+1))
/(2*n+1)}

Is there a way to prove it?

Answer 1

Let us rearrange the expression a bit, factoring terms not depending on $k$:

$$ T(n,m) = \frac{(n+m+1)!}{n!(m+1)!}\sum_{k=0}^{m}\binom{n}{k}\binom{n+m-k}{m-k} \frac{(-1)^k}{2k+1}. \tag{1} $$

If the term $\frac{(-1)^k}{2k+1}$ were not there, the RHS of $(1)$ would be given by Vandermonde's identity.

Anyway

$$ T(n,m) = \binom{n+m+1}{n}\int_{0}^{1}\sum_{k=0}^{m}\binom{n}{k}(-x^2)^k \binom{n+(m-k)}{(m-k)}\,dx \tag{2}$$

where the innermost sum equals the coefficient of $z^m$ in the product between

$$ \sum_{k\geq 0}\binom{n}{k}(-zx^2)^k = (1-zx^2)^n\quad\text{and}\quad \sum_{k\geq 0}\binom{n+k}{k}z^k = \frac{1}{(1-z)^{n+1}}.\tag{3}$$

It follows that

$$\begin{eqnarray*} T(n,m)&=&\binom{n+m+1}{n}\int_{0}^{1}\text{Res}_{z=0}\left(\frac{(1-zx^2)^n}{z^{m+1}(1-z)^{n+1}}\right)\,dx\\&=&\binom{n+m+1}{n}\int_{0}^{1}\frac{1}{2\sqrt{u}}\,\text{Res}_{z=0}\left(\frac{(1-zu)^n}{z^{m+1}(1-z)^{n+1}}\right)\,du\\&=&\binom{n+m+1}{n}\int_{0}^{1}\frac{u^{m+n+1}}{2u\sqrt{u}}\,\text{Res}_{z=0}\left(\frac{(1-z)^n}{z^{m+1}(u-z)^{n+1}}\right)\,du\\&=&-\binom{n+m+1}{n}\int_{0}^{1}\frac{u^{m+n+1}}{2\sqrt{u}}\,\text{Res}_{z=u}\left(\frac{(1-z)^n}{z^{m+1}(u-z)^{n+1}}\right)\,du\\&=&\binom{n+m+1}{n}\int_{0}^{1}\frac{u^{m+n+1}}{2\sqrt{u}}\,\frac{1}{n!}\cdot\left.\frac{d^n}{dz^n}\frac{(1-z)^n}{z^{m+1}}\right|_{z=u}\,du.\tag{4}\end{eqnarray*}$$

Integration by parts and Euler's Beta function readily prove the statement.

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Maybe there is some shortcut, since $\sum_{k=0}^{m}\binom{\color{red}{m}}{k}\frac{(-1)^k}{2k+1}=\frac{4^m}{(2m+1)\binom{2m}{m}}$ is a direct consequence of the binomial transform, and these coefficients are related to the Maclaurin expansion of $\arcsin^2(x)$ or $\frac{\arcsin(x)}{\sqrt{1-x^2}}$ (this is the reason behind my initial comment).

Answer 2

I was curious for some time how to solve this in elementary methods, without hypergeometric functions, gamma/beta functions, residues, etc, but with some snake oil. It is a bit more involved, but here goes:

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The sum that we want to evaluate: $$ \boxed{\sum\limits_{k} \binom{n+m+1}{m+k+1} \binom{n+m-k}{n} \binom{m+k+1}{k} \frac{(-1)^k}{2k+1}} $$ Note that $$ \binom{n+m+1}{m+k+1} \binom{m+k+1}{k} = \binom{n+m+1}{k, m+1,n-k} = \binom{n+m+1}{m+1} \binom{n}{k}, $$ so we can factor out $\binom{n+m+1}{m+1}$ in a way that only $\binom{n}{k}$ remains in the sum: $$ \binom{n+m+1}{m+1} \sum\limits_{k} \binom{n}{k} \binom{n+m-k}{n} \frac{(-1)^k}{2k+1} $$ We now use $\binom{n+m-k}{n} = [x^n](1+x)^{n+m-k}$ to rewrite the sum as $$ [x^n] (1+x)^{m} \sum\limits_k \binom{n}{k} \frac{(-1)^k}{2k+1} (1+x)^{n-k} $$

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It is annoying to deal with $2k+1$, so we use a substitution similar to that of Jack D'Aurizio: $$ \frac{1}{2k+1} = \int\limits_0^1 y^{2k} dy, $$ which turns the expression into $$ [x^n](1+x)^m \int\limits_0^1 \sum\limits_k \binom{n}{k} (-y^2)^k (1+x)^{n-k} dy, $$ which collapses into $$ S_{n,m}=\int\limits_0^1 [x^n](1+x)^m (1+x-y^2)^n dy. $$

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We integrate it by parts with $f(y) = [x^n](1+x)^m (1+x-y^2)^n$ to reduce the degree of $(1+x-y^2)^n$: $$ S_{n,m} = [yf(y)]\big|_0^1 - \int\limits_0^1 y f'(y) dy $$ For the boundary term, we have $0\cdot f(0)=0$ and $1 \cdot f(1) = [x^n](1+x)^m x^n = 1$, so $$ S_{n,m} = 1 +2n \int\limits_0^1 [x^n](1+x)^m y^2 (1+x-y^2)^{n-1} dy $$ We use $y^2 = (1+x) - (1+x-y^2)$ to further turn this into $$ (1+2n)S_{n,m} = 1 + 2n \int\limits_0^1 [x^n] (1+x)^{m+1} (1+x-y^2)^{n-1} dy $$

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The integral is similar to $S_{n-1,m+1}$, except the mismatch between $[x^n]$ and $(1+x-y^2)^{n-1}$. To mitigate it, we expand $$ (1+x)^{m+1} = (1+x)^m + x(1+x)^m = 1+x\sum\limits_{k=0}^m (1+x)^{m-k} $$ Plugging it into the integral, and using $[x^n](1+x-y^2)^{n-1} = 0$, we get $$ (1+2n) S_{n,m} = 1+2n\sum\limits_{k=0}^m S_{n-1,m-k} $$

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We can get it into finite form by subtracting the expansion of $S_{n,m-1}$: $$ (1+2n)(S_{n,m} - S_{n,m-1}) = 2n S_{n-1,m} $$ On the other hand, we can use the symmetry $S_{n,m} = S_{m,n}$ to turn it into $$ (1+2m)(S_{n,m}-S_{n-1,m}) = 2m S_{n,m-1} $$ We can eliminate $S_{n-1,m}$ from both identities to get $$ \frac{2n+1}{2n}(S_{n,m}-S_{n,m-1}) = S_{n,m} - \frac{2m}{2m+1} S_{n,m-1} $$ This further simplifies into $$ \frac{1}{2n} S_{n,m} = \left(\frac{2n+1}{2n}-\frac{2m}{2m+1}\right) S_{n,m-1} = \frac{2n+2m+1}{2n(2m+1)} S_{n,m-1}, $$ which finally turns into $$ S_{n,m} = \frac{2n+2m+1}{2m+1} S_{n,m-1} = \frac{(2n+3) \cdot (2n+5) \cdot \ldots \cdot (2n+2m+1)}{3\cdot 5 \cdot \ldots \cdot(2m+1)} = \frac{(2n+2m+1)!!}{(2n+1)!! (2m+1)!!} $$

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We can now re-express it as $$ S_{n,m}=\frac{(2n+2m+2)!}{(2n+2)!(2m+2)!} \frac{2^{n+1}(n+1)!2^{m+1}(m+1)!}{2^{n+m+1}(n+m+1)!} $$ and multiply it with $\binom{n+m+1}{m+1} = \frac{(n+m+1)!}{n!(m+1)!}$ to finally get $$ \boxed{\frac{(2n+2m+2)!}{(2n+1)!(2m+2)!} = \frac{1}{2n+1}\binom{2(n+m+1)}{2(m+1)}} $$

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Note: You may notice that Jack D'Aurizio uses the other substitution: $$ \binom{n+m-k}{m-k} = [x^{m}] x^k \frac{1}{(1-x)^{n+1}} $$ This leads to $$ [x^m]\frac{1}{(1-x)^{n+1}} \int\limits_0^1 \sum\limits_k \binom{n}{k} (-xy^2)^k dy $$ So, it instead collapses to $$ \int\limits_0^1 [x^m] \frac{(1-xy^2)^n}{(1-x)^{n+1}}dy, $$ which is then computed with the technique from his answer.