Proving an interesting binomial sum

Question

I am trying to prove the binomial identity $$\sum_{k = 0}^n \frac{(-1)^k}{2k+1} \binom{2n - 2k}{n-k} \binom{n - k}{k} = \frac{4^n}{n+1},$$ without using induction. Here $n$ is a non-negative integer. The factor of $2k + 1$ in the denominator suggests one try to integrate $x^{2k}$ between the limits of $0$ and $1$. To do this one would need to know what $$\sum_{k = 0}^n (-1)^k \binom{2n - 2k}{n - k} \binom{n - k}{k} x^{2k},$$ is in closed-form. It seems to involve (according to Mathematica) some hypergeometric function, the integral of which is by no means obvious.

I have also tried coefficient extraction methods $[z^n]$, but again the factor of $2k + 1$ in the denominator makes this problematic (at least for me).

So my question is, given the simple form of the expression on the right of this identity, is there a relatively easy way to prove this identity? Perhaps there is some simple trick I am missing?

Answer 1

That's simple if you recognize Legendre polynomials hidden in such sum. We have

$$ P_n(x) = \frac{1}{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}\binom{2n-2k}{n}x^{n-2k}\stackrel{\text{Rodrigues}}{=}\frac{1}{2^n n!}\cdot\frac{d^n}{dx^n}(x^2-1)^n \tag{1}$$ while

$$ S(n)=\sum_{k=0}^{\lfloor{n/2}\rfloor}\frac{(-1)^k}{2k+1}\binom{2n-2k}{n-k}\binom{n-k}{k}=\sum_{k=0}^{\lfloor{n/2}\rfloor}\frac{(-1)^k}{2k+1}\binom{n}{k}\binom{2n-2k}{n}\tag{2}$$ so

$$ S(n) = 2^n\int_{0}^{1}x^n P_n(1/x)\,dx =2^n \int_{1}^{+\infty}\frac{P_n(x)}{x^{n+2}}\,dx.\tag{3}$$ This allows to recover the generating function for $S(n)$ from the generating function of $P_n(x)$.
By multiplying both sides of $(3)$ by $z^n$ and summing over $n\geq 0$ we have $$\begin{eqnarray*} \sum_{n\geq 0} S(n) z^n &=& \int_{1}^{+\infty}\frac{dx}{x \sqrt{(1-4z)x^2+4z^2}}\\&=&\frac{1}{2z}\operatorname{arcsinh}{\frac{2z}{\sqrt{1-4z}}}\\&=&\frac{1}{2z}\operatorname{arctanh}\left(\frac{2z}{1-2z}\right)=\frac{-\ln(1-4z)}{4z}\end{eqnarray*}\tag{4}$$ and the claim readily follows from the Maclaurin series of the RHS of $(4)$.

Answer 2

Theorem: If $f$ is a polynomial of degree $n$ or less, $$ \sum_{k=0}^n(-1)^k\binom{n}{k}\frac{f(k)}{x+k}=\frac{n!f(-x)}{x(x+1)(x+2)\cdots(x+n)}\tag1 $$ Proof: Use the Heaviside Method for Partial Fractions: to get the coefficient for $\frac1{x+k}$ multiply by $x+k$ and set $x=-k$: $$ \frac{n!f(k)}{\underbrace{(-k)(-k+1)(-k+2)\cdots(-1)}_{(-1)^kk!}\underbrace{(1)\cdots(-k+n)}_{(n-k)!}}=(-1)^k\binom{n}{k}f(k)\tag2 $$ $\large\square$

We also have $$ \begin{align} \binom{2n-2k}{n-k}\binom{n-k}{k} &=\binom{2n-2k}{n-k}\binom{n-k}{n-2k}\tag{3a}\\ &=\binom{2n-2k}{n-2k}\binom{n}{k}\tag{3b}\\ &=\binom{2n-2k}{n}\binom{n}{k}\tag{3c}\\ \end{align} $$ Explanation:
$\text{(3a):}$ $\binom{a}{b}=\binom{a}{a-b}$ (symmetry of Pascal's Triangle)
$\text{(3b):}$ $\binom{a}{b}\binom{b}{c}=\binom{a}{c}\binom{a-c}{b-c}$ (expand as ratios of factorials)
$\text{(3c):}$ $\binom{a}{b}=\binom{a}{a-b}$ (symmetry of Pascal's Triangle)

Since $f(x)=\frac12\binom{2n-2x}{n}$ is a polynomial of degree $n$, we can apply $(1)$ with $x=\frac12$: $$ \begin{align} \sum_{k=0}^n\frac{(-1)^k}{\color{#C00}{2k+1}}\color{#090}{\binom{2n-2k}{n-k}\binom{n-k}{k}} &=\sum_{k=0}^n\frac{(-1)^k}{\color{#C00}{k+\frac12}}\color{#C00}{\frac12}\color{#090}{\binom{2n-2k}{n}\binom{n}{k}}\tag{4a}\\ &=\frac{n!\frac12\binom{2n+1}{n}}{\frac12\frac32\frac52\cdots\frac{2n+1}2}\tag{4b}\\ &=\frac{n!\binom{2n+1}{n}}{\frac{(2n+1)!}{4^nn!}}\tag{4c}\\[3pt] &=\frac{4^n}{n+1}\tag{4d} \end{align} $$ Explanation:
$\text{(4a):}$ apply $(3)$
$\text{(4b):}$ apply $(1)$ with $x=\frac12$ and $f(x)=\frac12\binom{2n-2x}{n}$
$\text{(4c):}$ cancel the $\frac12$ in the numerator and denominator
$\phantom{\text{(4c):}}$ and write the denominator in terms of factorials
$\text{(4d):}$ simplify

Answer 3

Here we use a technique that can be found in the classic Integral Representation and the Computation of Combinatorial Sums by G. P. Egorychev. In order to do so we write the binomial coefficient $\binom{n}{k}$ as residue of a meromorphic function, namely \begin{align*} \color{blue}{\binom{n}{k}}&=\frac{n!}{k!(n-k)!}=(-1)^{n-k}n!\left(\prod_{q=0}^{k-1}\frac{1}{k-q}\right) \left(\prod_{q=k+1}^n\frac{1}{k-q}\right)\\ &\,\,\color{blue}{=(-1)^{n-k}n!\operatorname{res}_{z=k}\prod_{q=0}^n\frac{1}{z-q}}\tag{1.1}\\ \binom{n}{k}h(k)&=(-1)^{n-k}n!\operatorname{res}_{z=k}\left(h(z)\prod_{q=0}^n\frac{1}{z-q}\right)\tag{1.2} \end{align*} The expression (1.2) is just a slightly more general variation of (1.1) which will be conveniently used in the following.

We obtain \begin{align*} \color{blue}{\sum_{k=0}^n}&\color{blue}{\frac{(-1)^k}{2k+1}\binom{2n-2k}{n-k}\binom{n-k}{k}}\\ &=\sum_{k=0}^n\frac{(-1)^k}{2k+1}\,\frac{(2n-2k)!}{(n-k)!}\,\frac{1}{k!(n-k)!}\tag{2.1}\\ &=\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{2k+1}\,\binom{2n-2k}{n}\\ &=\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{2k+1}\,\frac{1}{n!}\prod_{j=0}^{n-1}\left(2n-2k-j\right)\tag{2.2}\\ &=\sum_{k=0}^n\operatorname{res}_{z=k}\left(\frac{(-1)^n}{2z+1} \prod_{j=0}^{n-1}(2n-2z-j)\prod_{q=0}^n\frac{1}{z-q}\right)\tag{$\to\ $1.2}\\ &=\frac{(-1)^n}{2}\sum_{k=0}^n\operatorname{res}_{z=k}\underbrace{\left(\frac{1}{z+\frac{1}{2}} \prod_{j=0}^{n-1}(2n-j-2z)\prod_{q=0}^n\frac{1}{z-q}\right)}_{f(z)}\\ &=\frac{(-1)^n}{2}\left(-\operatorname{res}_{z=-\frac{1}{2}}f(z)-\operatorname{res}_{z=\infty} f(z)\right)\tag{2.3}\\ &\,\,\color{blue}{=\frac{(-1)^n}{2}\left(-\operatorname{res}_{z=-\frac{1}{2}}f(z)+\operatorname{res}_{z=0} \left(\frac{1}{z^2}f\left(\frac{1}{z}\right)\right)\right)}\tag{2.4}\\ \end{align*}

Comment:

• In (2.1) we use $\binom{p}{q}=\frac{p!}{q!(p-q)!}$ and do some cancellation.

• In (2.2) we $\binom{p}{q}=\frac{1}{q!}p(p-1)\cdots(p-q+1)$.

• In (2.3) we use the sum of the residues of a meromorphic function at the poles $z=q, 0\leq q\leq n$ and $z=-\frac{1}{2}$ plus the residue at $\infty$ sum up to zero. This way we get rid of the sum and what is left are just two residues, the one at $z=-\frac{1}{2}$ and the one at $z=\infty$.

• In (2.4) we use the identity \begin{align*} \operatorname{res}_{z=\infty}f(z)=\operatorname{res}_{z=0}\left(-\frac{1}{z^2}f\left(\frac{1}{z}\right)\right) \end{align*} which transforms a residue at infinity to a residue at zero.

We now calculate the residues. Since we already know the result is the simple expression $\frac{4^n}{n+1}$ a cool aspect is, we might expect a lot of cancellations. We start with the residue at $z=-\frac{1}{2}$ and obtain \begin{align*} \color{blue}{\operatorname{res}_{z=-\frac{1}{2}}f(z)} &=\operatorname{res}_{z=-\frac{1}{2}}\left(\frac{1}{z+\frac{1}{2}}\prod_{j=0}^{n-1}(2n-j-2z)\prod_{q=0}^n\frac{1}{z-q}\right)\\ &=\operatorname{\lim}_{z\to -\frac{1}{2}}\left(\prod_{j=0}^{n-1}(2n-j-2z)\prod_{q=0}^n\frac{1}{z-q}\right)\\ &=\prod_{j=0}^{n-1}(2n-j+1)\prod_{q=0}^n\frac{1}{-\frac{1}{2}-q}\\ &=(-1)^{n+1}2^{n+1}\prod_{j=0}^{n-1}(2n-j+1)\prod_{q=0}^n\frac{1}{2q+1}\\ &=(-1)^{n+1}2^{n+1}\frac{(2n+1)!}{(n+1)!}\,\frac{1}{(2n+1)!!}\\ &=(-1)^{n+1}2^{n+1}\frac{(2n+1)!}{(n+1)!}\,\frac{2^nn!}{(2n+1)!}\\ &\,\,\color{blue}{=(-1)^{n+1}2^{2n+1}\frac{1}{n+1}}\tag{3} \end{align*} In a similar way we use (2.4) and find the residue of $f(z)$ at $z=\infty$ is zero. Putting (2.4) and (3) together we finally obtain \begin{align*} \frac{(-1)^n}{2}&\left(-\operatorname{res}_{z=-\frac{1}{2}}+\operatorname{res}_{z=0} \left(\frac{1}{z^2}f\left(\frac{1}{z}\right)\right)\right)\\ &=\frac{(-1)^n}{2}\left((-1)^{n}2^{2n+1}\frac{1}{n+1}+0\right)\\ &\,\,\color{blue}{=\frac{4^{n}}{n+1}} \end{align*} and the claim follows.