How was this series of binomial coefficients evaluated?

Question

I have less than surface level knowledge on both evaluating convolution sums and using binomial coefficients (I think this is stuff learned or heavily studied in Discrete math which I myself never took), and even less on evaluating a combination of these 2 concepts. That said, I want to see all of the many moving parts involved in evaluating the following convolution series:

$$\tag{1} J_n = \frac{\sqrt{\pi}}{2^{2n}}\sum_{k=0}^{n} (-1)^k \binom{2n}{k}\binom{4n-2k}{2n}(k-n)\frac{(n-k-\frac{1}{2})!}{(n-k+1)!}$$

In the procedure for evaluating the series, several steps unknown to me were performed (apparently including manipulating the binomial coefficients) to obtain:

$$\tag{2}J_n = -\frac{n}{2^{2n}}\sqrt{\pi}\binom{4n}{2n}\frac{(n-\frac{1}{2})!}{(n+1)!}S_n$$

where $S_n$ is defined by using the following recurrence relation: $$a_0 = 1\\a_{k+1} = -\frac{(n-k-1)(n-k+1)}{(2n-k-\frac{1}{2})(k+1)}a_k$$ to identify

$$a_k = \frac{(1-n)^{(k)} (-1-n)^{(k)}}{(\frac{1}{2}-2n)^{(k)} k!}$$ in $S_n = \sum_{k=0}^n a_k$, where I just learned rencently that the notation: $^{(k)}$, is referred to as a $\textit{rising factorial}$. My only real experience with creating recurrence relations was from creating power series solutions to ODEs and using boundary conditions to evaluate $c_k$ only for $k=0$ and $k=1$, I've never evaluated $c_k$ for any and all $k$ and I've never identified a recurrence relation outside of that context.

So, we finally end up having to evaluate $$\tag{3}S_n = \sum_{k=0}^n \frac{(1-n)^{(k)} (-1-n)^{(k)}}{(\frac{1}{2}-2n)^{(k)} k!} = _2F_1\Big(1-n,-1-n;\frac{1}{2}-2n;1\Big)$$

Where $_2F_1(a,b;c;z)$ is something called a $\textit{hypergeometric function}$, and I guess these have their own method of evaluation which I do not know about. By evaluation of the hypergeometric function, the series in (3) evaluates to $$S_n = \sqrt{\pi}\frac{(-\frac{1}{2}-2n)!}{(-\frac{3}{2}-n)!(\frac{1}{2}-n)!}$$

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The properties I know of for Binomial coefficients (so far from the book $\textit{Concrete Mathematics}$ which I obtained yesterday) are the following:

• symmetry: $\binom{n}{k} = \binom{n}{n-k}$
• factoring: $\binom{n}{k} = \frac{n}{k}\binom{n-1}{k-1}$
• addition formula: $\binom{n}{k} = \binom{n-1}{k}+ \binom{n-1}{k-1}$
• polynomial argument: $(n-k)\binom{n}{k} = (n-k)\binom{n}{n-k} = n\frac{n-k}{n-k}\binom{n-1}{(n-1)-k} = n\frac{(n-1)!}{(n-1-k)!(n-1-(n-1-k))!}= n\binom{n-1}{k}$
• addition formula (falling factorial): $\binom{n}{k} = \binom{n-1}{k}+ \binom{n-1}{k-1} = \frac{(n-1)_k}{k!} + \frac{(n-1)_{k-1}}{(k-1)!} = \frac{(n-1)_k (n-k)}{k!}+\frac{(n-1)_{k-1}k}{k!}=\frac{(n-1)_{k-1}n }{k!}=\frac{(n)_k}{k!}$
• general summation formula (from addition formula): $\binom{n}{k} = \sum_{k\le n}^n \binom{n+k}{k} = \binom{n}{0}+\binom{n+1}{1}+...+\binom{n+n}{n} = \binom{n+n+1}{n} = (2n+1)\binom{2n}{n}$

where we note in the above formula that we've changed indices from $k$ to $n$.

• negatives: $\binom{-1}{k} = (-1)^k, \binom{-1}{-1-k} = (-1)^{-1-k}$

For factorials, I know all of the basic ones like $n! = n(n-1)!$, and some of the properties from the wikipedia on gamma functions:

• $\Gamma(1/2) = \sqrt{\pi}$
• $\Gamma(1/2+n) = \binom{n-1/2}{n}n!\sqrt{\pi}$
• $\Gamma (1/2-n) = \frac{\sqrt{\pi}}{\binom{-1/2}{n}n!}$

And that's about all I know for likely relevant properties. Surely other more advanced properties are applied throughout this problem starting with getting from (1) to (2), and I haven't even touched on deriving recurrence relations in the latter portions of this derivation.

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So, I want to see 1) what happened to $\binom{2n}{k}\binom{4n-2k}{2n}$ and how we end up with $\binom{4n}{2n}$, 2) the process by which we derive to a recurrence relation (I'm unfamiliar with the procedure for obtaining one in this context, and I have barely a surface level knowledge of recurrence relations), 3) how we derived $a_k$ from the recurrence relation (I don't think I've ever done this in any of my formal education from calculus to ODEs, but if this is trivially easy for this problem, any link showing and explaining procedure for how the recurrence relation is used to obtain this would suffice), and 4) how to evaluate the hypergeometric function (I have no knowledge of hypergeometric functions).

That's quite a lot, but as I said at the beginning, there are a lot of moving parts involved in moving from (1) to (2) which involve more than my less than surface level knowledge in the relevant subjects that this derivation demands greater proficiency in.

Answer 1

(You missed a factor of $\sqrt\pi$ in your initial $J_n$ which is included in the rest - I include it here.) The main goal of this is to get in the form of a hypergeometric function to be able to use Gauss's summation theorem. We start by making the first term "1". To do this we just divide out by the $k=0$ term which is:

$$ \frac{1}{2^{2n}} \sqrt\pi\binom{2n}{0}\binom{4n}{2n}(-n)\frac{(n-\frac{1}{2})!}{(n+1)!} = -\frac{n}{2^{2n}}\sqrt\pi\binom{4n}{2n}\frac{(n-\frac{1}{2})!}{(n+1)!} $$

Then we calculate the ratio between successive terms. This makes it easy to relate to raising factorials, as used in hypergeometric functions. Some people will be able to see the result without taking the ratio, but I find it easier this way to see nice cancellation between terms. You need to be careful, because the extra factors always come from the higher factorial, which is sometimes in the numerator and sometimes in the denominator:

\begin{align} \frac{a_{k+1}}{a_k} &= \frac{(-1)^{k+1} \binom{2n}{k+1} \binom{4n-2(k+1)}{2n}(k+1-n)\frac{(n-(k+1)-\frac12)!}{n-(k+1)+1)!}} {(-1)^k \binom{2n}{k} \binom{4n-2k}{2n}(k-n)\frac{(n-k-\frac12)!}{(n-k+1)!}}\\ &= -\frac{2n-k}{k+1} \frac{(2n-2k)(2n-2k-1)}{(4n-2k)(4n-2k-1)}\frac{k+1-n}{k-n}\frac{n-k+1}{n-k-\frac12}\\ &= -\frac{(n-k-1)(n-k+1)}{(2n-k-\frac12)(k+1)}\\ &= \frac{(1-n+k)(-1-n+k)}{(\frac12-2n+k)(k+1)} \end{align}

In the second last step we can directly see lowering factorials, with the $k+1$ giving a normal factorial. They are lowering because of the $-k$, but in the last step above, I took negatives to go straight to rising factorials:

$$a_k = \frac{(1-n)^{(k)}(-1-n)^{(k)}}{(\frac12-2n)^{(k)}k!}$$

The definition of the hypergeometric function gives us:

$$ {}_2F_1(1-n,-1-n;\tfrac12-2n;1) = \sum_{k=0}^n \frac{(1-n)^{(k)}(-1-n)^{(k)}}{(\frac12-2n)^{(k)}k!}$$

Then Gauss's summation theorem gives us

$$ {}_2F_1(1-n,-1-n;\tfrac12-2n;1) = \frac{\sqrt\pi\Gamma(\frac12-2n)}{\Gamma(-\frac12-n)\Gamma(\frac32-n)}$$

There is an alternative way to do this, by simplifying first, using the Legendre duplication formula $\Gamma(z)\Gamma(z+\frac12)=2^{1-2z}\Gamma(2z)$:

\begin{align} \binom{2n}{k}\binom{4n-2k}{2n}(k-n)\frac{\Gamma(n-k+\frac{1}{2})}{\Gamma(n-k+2)} &= \frac{k-n}{k!(2n-k)!}\frac{(4n-2k)!}{(2n-2k)!} \frac{\Gamma(n-k+\frac{1}{2})}{\Gamma(n-k+2)} \\ &= -\frac1{k!\Gamma(2n-k)}\frac{\Gamma(4n-2k)}{\Gamma(2n-2k)} \frac{\Gamma(n-k+\frac{1}{2})}{\Gamma(n-k+2)} \\ &= -2^{2n}\frac1{k!\Gamma(2n-k)}\frac{\Gamma(2n-k)\Gamma(2n-k+\frac12)}{\Gamma(n-k)\Gamma(n-k+\frac12)} \frac{\Gamma(n-k+\frac{1}{2})}{\Gamma(n-k+2)} \\ &= -2^{2n}\frac1{k!}\frac{\Gamma(2n-k+\frac12)}{\Gamma(n-k)\Gamma(n-k+2)} \end{align} So, again dividing out the $k=0$ factor, and using $(x)_n = \Gamma(x+1)/\Gamma(x-n+1)$, \begin{align} a_k &= (-1)^k\frac1{k!}\frac{\Gamma(2n-k+\frac12)/\Gamma(2n+\frac12)}{[\Gamma(n-k)/\Gamma(n)][\Gamma(n-k+2)/\Gamma(n+2)]} \\ a_k &= (-1)^k\frac{(n-1)_k(n+1)_k}{(2n-\frac12)_k k!} \\ &= \frac{(1-n)^{(k)}(-1-n)^{(k)}}{(\frac12-2n)^{(k)} k!} \end{align} as before, using $(x)_k=(-1)^k(-x)^{(k)}$.

Answer 2

Supposing we are interested in

$$J_n = \frac{1}{2^{2n}} \sum_{k=0}^n (-1)^k {2n\choose k} {4n-2k\choose 2n} (n-k) \frac{\Gamma(n-k+1/2)}{(n-k+1)!}.$$

Consulting Wikipedia on the Gamma function we can re-write the RHS as

$$\frac{1}{2^{2n}} \sum_{k=0}^n (-1)^k {2n\choose k} {4n-2k\choose 2n} \frac{n-k}{(n-k+1)!} \sqrt\pi \frac{(2n-2k)!}{2^{2n-2k} (n-k)!} \\ = \frac{\sqrt\pi}{2^{4n}} \sum_{k=0}^n (-1)^k {2n\choose k} {4n-2k\choose 2n} {2n-2k\choose n-k+1} 2^{2k}.$$

Doing a complete cancellation on the binomial coefficients we find

$$\frac{(4n-2k)!}{k! \times (2n-k)! \times (n-k+1)! \times (n-k-1)!} \\ = {n+1\choose k} \frac{(4n-2k)!}{(2n-k)! \times (n+1)! \times (n-k-1)!} \\ = {n+1\choose k} {2n-k\choose n+1} {4n-2k\choose 2n-k}.$$

Note that here the value $k=n+1$ will produce zero from the middle binomial coefficient so we may include it in the sum, getting

$$\frac{\sqrt\pi}{2^{4n}} [z^{n+1}] (1+z)^{2n} \sum_{k=0}^{n+1} {n+1\choose k} \frac{1}{(1+z)^k} {4n-2k\choose 2n-k} (-1)^k 2^{2k} \\ = \sqrt\pi [z^{n+1}] (1+z)^{2n} \sum_{k=0}^{n+1} {n+1\choose k} \frac{(-1)^k}{(1+z)^k} \frac{1}{2^{4n-2k}} [w^{2n-k}] \frac{1}{\sqrt{1-4w}} \\ = \sqrt\pi [z^{n+1}] (1+z)^{2n} [w^{2n}] \frac{1}{\sqrt{1-w}} \sum_{k=0}^{n+1} {n+1\choose k} \frac{(-1)^k}{(1+z)^k} w^k \\ = \sqrt\pi [z^{n+1}] (1+z)^{2n} [w^{2n}] \frac{1}{\sqrt{1-w}} \left[1-\frac{w}{1+z}\right]^{n+1} \\ = \sqrt\pi [z^{n+1}] (1+z)^{n-1} [w^{2n}] \frac{1}{\sqrt{1-w}} \left[1-w+z\right]^{n+1}.$$

The contribution from $z$ is

$$\;\underset{z}{\mathrm{res}}\; \frac{1}{z^{n+2}} (1+z)^{n-1} [1-w+z]^{n+1}.$$

Now put $z/(1+z)=u$ so that $z=u/(1-u)$ and $dz = 1/(1-u)^2\; du$ to get

$$\;\underset{u}{\mathrm{res}}\; \frac{1}{u^{n+2}} (1-u)^3 [1-w+u/(1-u)]^{n+1} \frac{1}{(1-u)^2} \\ = \;\underset{u}{\mathrm{res}}\; \frac{1}{u^{n+2}} \frac{1}{(1-u)^n} [1-w+uw]^{n+1}.$$

Here the residue at infinity is zero and we may evaluate using minus the residue at $u=1$ which requires the Leibniz rule:

$$\frac{1}{(n-1)!} \left( \frac{1}{u^{n+2}} [1-w+uw]^{n+1} \right)^{(n-1)} = \frac{1}{(n-1)!} \\ \times \sum_{q=0}^{n-1} {n-1\choose q} \frac{(-1)^q (n+2)^{\overline{q}}}{u^{n+2+q}} (n+1)^{\underline{n-1-q}} [1-w+uw]^{n+1-(n-1-q)} w^{n-1-q}.$$

Now put $u=1$ to get

$$(-1)^{n+1} \sum_{q=0}^{n-1} (-1)^q {n+1+q\choose q} {n+1\choose q+2} w^{n-1-q}.$$

With the extractor in $w$,

$$(-1)^{n+1} \sqrt\pi \sum_{q=0}^{n-1} (-1)^q {n+1+q\choose q} {n+1\choose q+2} [w^{n+1+q}] \frac{1}{\sqrt{1-w}} \\ = \sqrt\pi \sum_{q=0}^{n-1} {n+1+q\choose q} {n+1\choose q+2} {-1/2\choose n+1+q}.$$

Now observe that (this also gives the right answer when $q=0$)

$${n+1+q\choose q} {-1/2\choose n+1+q} = \frac{1}{q!\times (n+1)!} \prod_{p=0}^{n+q} (-1/2-p) \\ = \frac{1}{q!\times (n+1)!} \prod_{p=0}^n (-1/2-p) \prod_{p=n+1}^{n+q} (-1/2-p) \\ = \frac{1}{q!} {-1/2\choose n+1} \prod_{p=0}^{q-1} (-3/2-n-p) = {-1/2\choose n+1} {-3/2-n\choose q}.$$

We therefore have for the target sum

$$\sqrt\pi {-1/2\choose n+1} \sum_{q=0}^{n-1} {n+1\choose q+2} {-3/2-n\choose q}.$$

At last apply Chu-Vandermonde to get

$$\bbox[5px,border:2px solid #00A000]{ \sqrt\pi {-1/2\choose n+1} {-1/2\choose n-1}.}$$

Answer 3

Similar to Marko Riedel, I will use $n-k$ instead of $k-n$ to ensure that $J_n$ is non-negative: $$ J_n = \frac{\sqrt{\pi}}{4^{n}}\sum_{k=0}^{n} (-1)^k \binom{2n}{k}\binom{4n-2k}{2n}(n-k)\frac{(n-k-\frac{1}{2})!}{(n-k+1)!} $$ Note that $(-\frac{1}{2})! = \Gamma(\frac{1}{2})=\sqrt\pi$ (see this for geometric interpretation), so $$ \small\left(n-\frac{1}{2}\right)! = \left(-\frac{1}{2}\right)! \prod\limits_{t=1}^{n} \left(t-\frac{1}{2}\right) = \sqrt{\pi} \frac{(2n-1)!!}{2^{n}} = \sqrt{\pi} \frac{(2n)!}{4^{n}n!} $$ Plugging it back into $J_n$ and simplifying the expression, we get $$ J_n = \frac{\pi}{16^{n}} \sum_{k} (-4)^{k} \binom{2n}{k}\binom{4n-2k}{2n}\binom{2n-2k}{n-k+1} $$ Let's substitute $a=n-1$ and $b=n+1$, as it will be easier to analyze: $$ J_n = \frac{\pi}{16^n} \sum\limits_k (-4)^k \binom{a+b}{k} \binom{2(a+b-k)}{a+b}\binom{a+b-2k}{a-k} $$ Note that the product of binomials is, in fact, a multinomial coefficient that we can rewrite as $$ \binom{2(a+b-k)}{k,a+b-k,a-k,b-k} = \binom{2(a+b-k)}{a+b-k}\binom{a+b-k}{a} \binom{a}{k} $$ Making a substitution $k \mapsto a+b-k$, we turn the sum into $$ \sum\limits_k (-4)^{a+b-k} \binom{2k}{k} \binom{k}{a} \binom{a}{k-b} $$ This form is much nicer, as we can now sum it up with a bivariate generating function: $$ G(x, y) = \sum\limits_{a,b,k} (-4)^{a+b-k} \binom{2k}{k} \binom{k}{a} \binom{a}{k-b} x^a y^b $$ Then, we're interested in the value of $[x^a y^b] G(x, y)$. To find it, we rewrite $$ G(x, y) = \sum\limits_k \frac{1}{(-4)^k} \binom{2k}{k} \sum\limits_a \binom{k}{a} (-4x)^a \sum\limits_{b}\binom{a}{k-b} (-4y)^b $$ The inner-most sum collapses into $$ \sum\limits_{b} \binom{a}{b} (-4y)^{k-b} = (-4y)^{k-a}(1-4y)^a $$ Then, the next sum turns into $$ (-4)^k\sum\limits_a \binom{k}{a} x^a (1-4y)^a y^{k-a} = (-4)^k(x+y-4xy)^k $$ Finally, using $\frac{1}{\sqrt{1-4x}} = \sum\limits_k \binom{2k}{k} x^k$, we turn the outer sum into $$ \sum\limits_k \binom{2k}{k} (x+y-4xy)^k = \frac{1}{\sqrt{1-4(x+y-4xy)}}=\frac{1}{\sqrt{1-4x}}\frac{1}{\sqrt{1-4y}} $$ This means that $[x^a y^b] G(x, y) = \binom{2a}{a} \binom{2b}{b}$, therefore $$ \boxed{J_n = \frac{\pi}{16^n} \binom{2(n-1)}{n-1} \binom{2(n+1)}{n+1}} $$ Which is another version of the answer that Marko Riedel obtained.

P.S. I would imagine that the identity $$ \sum\limits_k (-4)^{a+b-k} \binom{2k}{k} \binom{k}{a} \binom{a}{k-b}=\binom{2a}{a} \binom{2b}{b} $$ also has a combinatorial meaning, but it is not obvious to me.