# 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:

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

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

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 $0 < Re\,s < 2n$ $$[2]\,\,\,\,\,\sum_{k=1}^n (-1)^{k+1} \binom{2n}{n+k} k^s = 2^{2n-s}\frac{s!}{\pi}\, \sin(\pi s/2) \int_0^\infty \frac{dt}{t^{s+1}}\, \sin^{2n}t.$$ 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 $$[3]\,\,\,\,\,(2\sin t)^{2n}= 2\sum_{k=1}^n (-1)^k\binom{2n}{n+k}\cos(2\, k\, t) + \binom{2n}{n}.$$ Insert Left Hand Side (LHS) of previous equation into integral on RHS of eq [2]. 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 [2]. Analytic continuation permits the $s$ to be extended to the stated range. The proof of [2] 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. [1]. $$s!\int_0^\infty \sin^{2n}\!t\frac{dt}{t^{s+1}}= \int_0^\infty dt \sin^{2n}t\int_0^\infty\exp{(-xt)}x^s dx = \int_0^\infty dx \, x^s \int_0^\infty dt\, e^{-xt}\sin^{2n}\!t$$

Now we'll show $$[4]\,\,\,\, J := \int_0^\infty dt\, e^{-xt}\big(2\,\sin{t}\big)^{2n} = (2n)! \frac{i}{2} \frac{\Gamma(ix/2)}{(n+ix/2)!} \frac{\Gamma(1-ix/2)}{(n-ix/2)!}$$ Again insert [3] 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 [4]. Use the $\Gamma$ reflection formula, rescale the integral and do some algebra and the proof of eq. [1] 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