# Alternate proof for weighted alternating shifted central binomial sum relation

In recent posts, MSE 2824529 and MSE 2825442, both initiated by user196574, I answered two asymptotic questions for $n \to \infty$ with the following identity:

$$\,\,\,\,\,\sum_{k=1}^n (-1)^{k+1} \binom{2n}{n+k} k^s = \binom{2n}{n} \sin(\pi s/2) \int_0^\infty \frac{dx \, \,x^s}{\sinh{\pi x}} \frac{n!^2}{(n+ix)!(n-ix)!}.$$

$\textit{To make this a legitimate question, I'm asking for a simpler proof or a generalization}$ $\textit{in which this formula is a special case. For scoring, Generalization > Simplicity.}$

An example of a generalization might be a Dirichlet character analog. So on the left there is squeezed in some sequence of (+1,-1,0) and on the RHS the hyperbolic trig function might get replaced with something more complicated, like a ratio of a linear combination of hyperbolic trig functions.

My proof, as requested from two contributors, will follow.

I was also asked for the motivation for such a formula. It wasn't for asymptotics, but instead to cobble a proof of the functional equation for Riemann's zeta function (at least formally) by using a hypergeometric identity. Since there are many rigorous proofs for the functional equation (and I don't publish anyway) I'd almost forgotten about it. If someone can generalize the formula as suggested above, it may be worth a revisit.

$\textbf{Proof}$

Establish the formula, valid for $0 < Re\,s < 2n$ $$\,\,\,\,\,\sum_{k=1}^n (-1)^{k+1} \binom{2n}{n+k} k^s = 2^{2n-s}\frac{s!}{\pi}\, \sin(\pi s/2) \int_0^\infty \frac{dt}{t^{s+1}}\, \sin^{2n}t.$$ The next few lines reproduce a well-known trig ID. By the binomial theorem $$\sum_{k=0}^{2n}\binom{2n}{k}(-x)^k = (x-1)^{2n}=\sum_{k=-n}^n\binom{2n}{k+n}(-x)^{k+n}$$ By splitting the sum at $k=-1$ and $k=1$ we have $$\sum_{k=1}^n (-1)^k\binom{2n}{n+k}\big(x^k + x^{-k} \big) = (-1)^n(x-1)^{2n}\,x^{-n} - \binom{2n}{n}.$$ Insert $x=\exp{(2it)}$ to get $$\,\,\,\,\,(2\sin t)^{2n}= 2\sum_{k=1}^n (-1)^k\binom{2n}{n+k}\cos(2\, k\, t) + \binom{2n}{n}.$$ Insert Left Hand Side (LHS) of previous equation into integral on RHS of eq . Also insert a regularizer $\exp{(-p\,t)}$ and we'll let $p \to 0$:

$$\int_0^\infty \frac{(2\sin{t})^{2n}}{t^{s+1}}dt = \lim_{p\to 0} \int_0^\infty dt \Big(2\sum_{k=1}^n (-1)^k\binom{2n}{n+k}\cos(2\, k\, t) + \binom{2n}{n} \Big)\frac{e^{-pt}}{t^{s+1}}.$$ $$= \lim_{p\to 0} 2\sum_{k=1}^n (-1)^k\binom{2n}{n+k}\Gamma(-s)\cos\big(s \arctan(2k/p)\big)(p^2+(2k)^2)^{s/2} + \binom{2n}{n}\Gamma(-s)p^{s}$$ where the Euler integral for the $\Gamma$ function has been used. The requirement $s>0$ and the assumption $s$ not an integer will mean the last term $\to$ 0. The last assumption will be relaxed upon analytic continuation. Taking the limit, e.g. $lim_{p\to 0} \arctan(2k/p)=\pi/2$ and using the reflection formula for the $\Gamma$ function allows us to write the result as eq . Analytic continuation permits the $s$ to be extended to the stated range. The proof of  is complete.

Now the $t^{-s-1}$ will be represented by the Euler integral to derive a double integral that will eventually give rise to the form in eq. . $$s!\int_0^\infty \sin^{2n}\!t\frac{dt}{t^{s+1}}= \int_0^\infty dt \sin^{2n}t\int_0^\infty\exp{(-xt)}x^s dx = \int_0^\infty dx \, x^s \int_0^\infty dt\, e^{-xt}\sin^{2n}\!t$$

Now we'll show $$\,\,\,\, J := \int_0^\infty dt\, e^{-xt}\big(2\,\sin{t}\big)^{2n} = (2n)! \frac{i}{2} \frac{\Gamma(ix/2)}{(n+ix/2)!} \frac{\Gamma(1-ix/2)}{(n-ix/2)!}$$ Again insert  into the LHS of 4, and again use the Euler integral to find $$J=2\sum_{k=1}^n (-1)^k\binom{2n}{n+k}\frac{x}{x^2+(2k)^2} + \binom{2n}{n}\frac{1}{x}$$ $$=\binom{2n}{n}\Big( 2\sum_{k=1}^n\frac{(-n)_k}{(n+1)_k}\frac{x}{x^2+(2k)^2} + \frac{1}{x} \Big)$$ where the expression in the last line uses the Pochhammer symbol. Now the following identity is known, e.g.,$\textit{Table of Series and Products},$ Hansen 6.6.57: $$\sum_{k=1}^n \frac{(-n)_k}{(n+1)_k}\frac{1}{k^2-a^2} = \frac{1}{2a^2} \Big(1-\frac{n!^2}{(1+a)_n\,(1-a)_n} \Big).$$ Use of it completes the proof of . Use the $\Gamma$ reflection formula, rescale the integral and do some algebra and the proof of eq.  is complete.

• Integrating (modulo prefactors) $$f(z)=\frac{z^s}{\sinh(\pi z)}\frac{1}{\Gamma(n+i z+1)\Gamma(n-i z+1)}$$ around a keyhole contour in the complex plane proofs this result pretty fast – tired Jun 22 '18 at 15:24
• @tired Give the question about a week, and if no generalization if forthcoming, I'll accept your comment, written as an answer. Checked your profile and saw your frustration with MSE. I'm about ready to quit. Very little feedback for thoughtful answers or questions, you have to search through 50 bad questions to find one good one, people who have rep have additional rep accrue easily (but newbies can't get it), user with rep editing my answer to obscure the fact that I corrected the OP's statement of problem (which made it a good problem indeed) etc. – skbmoore Jun 25 '18 at 16:35

## 1 Answer

Sorry to bump an old question, but I thought this would interest you.

In this paper with Divesh Aggarwal, Huck Bennett, and Sasha Golovnev, we ended up using this identity and a generalization of it to prove certain computational hardness results for lattice problems. See Section 7 there.

Here's the full statement: OP's statement corresponds to the case when $$k = 2\tau$$.

The proof uses the contour integration idea suggested by @tired.

Anyway, thanks for posting this! I doubt we would have discovered this identity without you!

• Thank you for including a call out in your paper. It really means a lot to me since I don't get to do that kind of research. Your generalization is very good, too. In another problem I asked for a different kind of generalization, b/c with formula  I was able to prove the function equation of the Riemann zeta function. – skbmoore May 4 '20 at 3:30