Proof of a new identity for the finite sum $\sum _{k=0}^{n-1} (-1)^k \sec \left(\frac{\pi\, k}{n}+\frac{\pi }{2 \ n}\right)$

Question

Problem statement:
Prove the following identity
\begin{align} \frac{\pi}{2\,\mathscr{n}} \sum _{k=0}^{\mathscr{n}-1} (-1)^k \sec \left(\frac{\pi\, k} {\mathscr{n}}\,+\,\tag{1}\label{eq:1} \frac{\pi }{2\,\mathscr{n}}\right)+\frac{\pi}{2} \sec \left(\frac{\pi\, \mathscr{n}}{2}\right)\, = \end{align}

\begin{align*} =\, \frac{\left(1-\frac{1}{\mathscr{n}}\right)}{(2 \pi )^{\mathscr{n}-1}}\,\, \mathcal{G_\texttt{1}(\mathscr{n})} \, + \, \frac{2}{(2 \pi )^{\mathscr{n}-1}}\,\,\mathcal{G_\texttt{2}(\mathscr{n})} \end{align*}
\begin{align*} =\, \frac{\left(1-\frac{1}{\mathscr{n}}\right)}{(2 \pi )^{\mathscr{n}-1}}\,\, % G_{\mathscr{n} + 2,\mathscr{n} + 2}^{\mathscr{n} + 2,\mathscr{n} + 1}\left( 1\left\vert \begin{array}{c} \frac{\frac{1-\mathscr{n}}{2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}},0,1\\ 0,0,\frac{\frac{1-\mathscr{n}}{2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}}% \end{array}% \right. \right)\,+ \end{align*}

\begin{align*} +\, \frac{2}{(2 \pi )^{\mathscr{n}-1}}\,\,% G_{\mathscr{n} + 2,\mathscr{n} + 2}^{\mathscr{n} + 2,\mathscr{n} + 1}\left( 1\left\vert \begin{array}{c} \frac{\frac{1-\mathscr{n}}{2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}},0,1\\ 0,0,\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 2}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n}{\mathscr{n}}% \end{array}% \right. \right) \end{align*}

by serial representations of the two Meijer G-functions on the right hand side. The finite sum on the left hand side converges for any even number $\mathscr{n}$.

The identity is a special case of a more general solution of the integral Vladimir Reshetnikov and can be derived starting from stocha - Summary (1. - 3.) - and solve the integral with the help of the Plancherel theorem, similar to Sasha. The more challenge is to prove the identity by a serial representation of the right hand side of \eqref{eq:1}. As a result new series representations of each Meijer G function follow, respectively.

Introduction:
For a more general discussion the representation of the Meijer G-functions through their generalizations - the Fox H - Functions - Mathai are needed:
\begin{align*} \mathcal{H_\texttt{1}(\mathscr{n})}\, = \mathcal{G_\texttt{1}(\mathscr{n})}\, = H_{3,3}^{3,2}\left( \tag{2}\label{eq:2}1\left\vert \begin{array}{c} \left(a_1,A_1\right) ,\left(a_2,A_2\right) ,\left(a_3,A_3\right) \\ \left(b_1,B_1\right) ,\left(b_2,B_2\right) ,\left(b_3,B_3\right) \end{array}\right. \right) \,\text{and}\qquad \end{align*} \begin{align*} \mathcal{H_\texttt{2}(\mathscr{n})}\, = \mathcal{G_\texttt{2}(\mathscr{n})}\, = H_{3,3}^{3,2}\left( 1\left\vert \begin{array}{c} \left(a_1,A_1\right) ,\left(a_2,A_2\right) ,\left(a_3,A_3\right) \\ \left(b_1,B_1\right) ,\left(b_2,B_2\right) ,\left(b_3 + 1,B_3\right) \end{array}\right. \right)\text{,} \end{align*}
where we used the following coefficients and indices:
\begin{align*} \left(a_1,A_1\right) = \left(\frac{1-\mathscr{n}}{2},\mathscr{n}\right),\, \left(a_n,A_n\right) = \left(a_2,A_2\right) = \left(0,1\right) \end{align*} \begin{align*} \left(a_p,A_p\right) = \left(a_{n+1},A_{n+1}\right) = \left(a_ 3,A_3\right) = \left(1,1\right) \end{align*}
\begin{align*} \left(b_1,B_1\right) = \left(0,1\right),\, \left(b_2,B_2\right) = \left(0,1\right) \end{align*} \begin{align*} \left(b_q,B_q\right) = \left(b_m,B_m\right) = \left(b_3,B_3\right) = \left(a_1,A_1\right) = \left(\frac{1-\mathscr{n}}{2},\mathscr{n}\right) \end{align*}
\begin{align*} m = 3,\, n = 2,\, p = 3,\, q = 3 \end{align*}
Motivation:

The identity is a special case of a more general solution of the integral Vladimir Reshetnikov

\begin{align} \int_{0}^\infty\operatorname{sech}(x)\operatorname{sech}(a\, x)\ dx\,=\,\left( 1-\frac{1}{\mathscr{n}}\right)\,\mathcal{H_\texttt{1}(\mathscr{a})}\,+\, \frac{2}{\mathscr{n}}\,\mathcal{H_\texttt{2}(\mathscr{a})}\tag{3}\label{eq:3}\\ \end{align}

\begin{align*} =\left( 1-\frac{1}{\mathscr{n}}\right)\,H_{3,3}^{3,2}\left( 1\left\vert \begin{array}{c} \left(a_1,A_1\right) ,\left(a_2,A_2\right) ,\left(a_3,A_3\right) \\ \left(b_1,B_1\right) ,\left(b_2,B_2\right) ,\left(b_3,B_3\right) \end{array}\right. \right)\,+ \end{align*}

\begin{align*} +\,\frac{2}{\mathscr{n}}\, H_{3,3}^{3,2}\left( 1\left\vert \begin{array}{c} \left(a_1,A_1\right) ,\left(a_2,A_2\right) ,\left(a_3,A_3\right) \\ \left(b_1,B_1\right) ,\left(b_2,B_2\right) ,\left(b_3 + 1,B_3\right) \end{array}\right. \right) \end{align*}

Besides of the identity \eqref{eq:1}, which is subject of this post, other interesting unknown identities arise from the discussion in stocha and especially the interesting answer Somos. An asymptotic expansion of $\mathcal{H_\texttt{1}(\mathscr{n})}$ and $\mathcal{H_\texttt{2}(\mathscr{n})}$ may also give more insight in the approximations of Claude Leibovici.

Moreover the study of the Mellin transformation of $\mathcal{H_\texttt{1}(\mathscr{n})}$ and $\mathcal{H_\texttt{2}(\mathscr{n})}$ is a valuable replenishment to the paper "Evaluation of a class of definite integrals" by M.L. Glasser (Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz.76,498-No.541(1975), 49-50.)

Since the argument of the Fox H-functions are equal to 1, we can use Mathai and write $\mathcal{H_\texttt{1}(\mathscr{n})}$ and $\mathcal{H_\texttt{2}(\mathscr{n})}$ in different forms:
\begin{align*} \mathcal{H_\texttt{1}(\mathscr{n})} = H_{3,3}^{3,2}\left( 1\left\vert \tag{4}\label{eq:4} \begin{array}{c} \left(a_p,A_p\right) \\ \left(b_q,B_q\right) \end{array}\right. \right) = H_{3,3}^{2,3}\left( 1\left\vert \begin{array}{c} \left(1-b_q,B_q\right) \\ \left(1-a_p,A_p\right) \end{array}\right. \right) = \frac{1}{k} \,H_{3,3}^{3,2}\left( 1\left\vert \begin{array}{c} \left(a_p,\frac{A_p}{k}\right) \\ \left(b_q,\frac{B_q}{k}\right) \end{array}\right. \right)\,\text{and}\qquad \end{align*}
\begin{align*} \mathcal{H_\texttt{2}(\mathscr{n})} = H_{3,3}^{3,2}\left( 1\left\vert \begin{array}{c} \left(a_p,A_p\right) \\ \left(b_q + 1,B_q\right) \end{array}\right. \right) = H_{3,3}^{2,3}\left( 1\left\vert \begin{array}{c} \left(1-b_q + 1,B_q\right) \\ \left(1-a_p,A_p\right) \end{array}\right. \right) \end{align*}

\begin{align*} = \frac{1}{k} \,H_{3,3}^{3,2}\left( 1\left\vert \begin{array}{c} \left(a_p,\frac{A_p}{k}\right) \\ \left(b_q + 1,\frac{B_q}{k}\right) \end{array}\right. \right) \end{align*}

where $k > 0$

Attempt at a solution:
For the prove we followed two ways: The first way is to express the Fox H - Functions $\mathcal{H_\texttt{1}(\mathscr{n})}$ and $\mathcal{H_\texttt{2}(\mathscr{n})}$ in serial representation Mathai, take advantage of the fact, that the poles of $\Gamma \left(1-a_j+s A_j\right)$ at least for $\mathcal{H_\texttt{1}(\mathscr{n})}$ are simple and receive an infinite sum:

\begin{align} \mathcal{H_\texttt{1}(\mathscr{n})} = (-1)^{\frac{\mathscr{n}}{2}} \,\pi \,\log\left(2\right) + \frac{1}{\mathscr{n}}\,\sum _{\nu =0}^{\infty } (-1)^{\nu } \,\frac{\Gamma\left(\frac{\mathscr{n} - 2\,\nu -1}{2\,\mathscr{n} }\right) \Gamma\left(1 + \nu\right) \Gamma\left(\frac{1 + \mathscr{n} + 2\,\nu }{2\, \mathscr{n} }\right)^2}{\nu ! \,\Gamma\left(\frac{1 + 3 \,\mathscr{n} + 2 \,\nu }{2\, \mathscr{n} }\right)} \tag{5}\label{eq:5} \end{align}

\begin{align*} = (-1)^{\frac{\mathscr{n}}{2}} \,\pi \,\log\left(2\right) + 2 \pi \, \sum _{\nu =0}^{\infty } \frac{(-1)^{\nu }\, \sec\left(\pi\,\frac{1+2\,\nu }{2\, \mathscr{n} }\right)}{1 + \mathscr{n} + 2 \,\nu } \end{align*}

It turned out, that the equivalent sum for $\mathcal{H_\texttt{2}(\mathscr{n})}$ in contrast do not converge. Therefore we use the series representations of the Meijer G function Mathematical Functions Site, which do converge:

\begin{align} \mathcal{G_\texttt{1}(\mathscr{n})} = \sum _{k=1}^{\mathscr{n}+1} \frac{\left(\left(\prod _{j=1,j\neq k}^{\mathscr{n}+1} \Gamma \left(a_k-a_j\right)\right) \prod _{j=1}^{\mathscr{n}+2} \Gamma\left(1+b_j-a_k\right)\right)}{\Gamma\left(1+a_{\mathscr{n}+2}-a_k\right)}\,_{\mathscr{n}+2}F_{\mathscr{n}+1}(\left\{1+b_{1}-a_k,\ldots ,1+b_{\mathscr{n}+2}-a_k\right\}, \tag{6}\label{eq:6} \end{align} \begin{align*} \left\{1+a_1-a_k,\ldots ,1+a_{k-1}-a_k,1+a_{k+1}-a_k,\ldots,1+a_{\mathscr{n}+2}-a_k\right\},-1)\\ \end{align*}

Note that $b_{1} = b_{2}=0$, $a_{\mathscr{n}+1} = 0$, $a_{\mathscr{n}+2} = 1$, $a_j = b_{j+2},\,j = 1,\ldots,n$ and

\begin{align} \frac{{\Gamma\left(1-a_{\mathscr{n}+1}\right)}^2}{\Gamma\left(2-a_{\mathscr{n}+1}\right)} = \frac{\Gamma\left(2-a_{\mathscr{n}+1}\right)}{{\left(a_{\mathscr{n}+1}-1\right)}^2}\tag{7}\label{eq:7}\\ \end{align}

\begin{align} \mathcal{G_\texttt{1}(\mathscr{n})} = \sum _{k=1}^{\mathscr{n}+1} \left(\left(\prod _{j=1,j\neq k}^{\mathscr{n}+1} \Gamma \left(a_k-a_j\right)\right) \prod _{j=1}^{\mathscr{n}} \Gamma\left(1+a_j-a_k\right)\right)\frac{\Gamma\left(2-a_{k}\right)}{{\left(a_{k}-1\right)}^2} \tag{8}\label{eq:8}\\ \end{align} \begin{align*} \,_{\mathscr{n}+2}F_{\mathscr{n}+1}(\left\{1-a_k,1-a_k,1+a_{1}-a_k,\ldots ,1+a_{\mathscr{n}}-a_k\right\}, \end{align*} \begin{align*} \left\{1+a_1-a_k,\ldots ,1+a_{k-1}-a_k,1+a_{k+1}-a_k,\ldots,1+a_{\mathscr{n}+2}-a_k\right\},-1)\\ \end{align*}

\begin{align} \mathcal{G_\texttt{1}(\mathscr{n})} = \sum _{k=1}^{\mathscr{n}+1} \left(\left(\prod _{j=1,j\neq k}^{\mathscr{n}+1} \Gamma \left(a_k-a_j\right)\right) \prod _{j=1}^{\mathscr{n}} \Gamma\left(1+a_j-a_k\right)\right)\frac{\Gamma\left(2-a_{k}\right)}{{\left(a_{k}-1\right)}^2} \tag{9}\label{eq:9} \end{align} \begin{align*} \,_{3}F_{2}(\left\{1,1-a_k,1-a_k\right\},\left\{1-a_k,2-a_k\right\},-1)\\ \end{align*}

\begin{align} \mathcal{G_\texttt{1}(\mathscr{n})} = \frac{1}{2}\,\sum _{k=1}^{\mathscr{n}+1} \left(\left(\prod _{j=1,j\neq k}^{\mathscr{n}+1} \Gamma \left(a_k-a_j\right)\right) \prod _{j=1}^{\mathscr{n}} \Gamma\left(1+a_j-a_k\right)\right)\frac{\Gamma\left(2-a_{k}\right)}{\left(a_{k}-1\right)}\,\left(\psi\left(\frac{1}{2}-\frac{a_k}{2}\right)-\psi\left(1-\frac{a_k}{2}\right)\right) \tag{10}\label{eq:10}\\ \end{align}

Note that $a_{\mathscr{n}+1} = 0$. For further simplifications we have to express the coefficients $a_k$ and $a_j$ as function of $\mathscr{n}$
\begin{align} a_j\left(\mathscr{n}\right) = \frac{1-\mathscr{n}}{2 \mathscr{n}}+\frac{j-1}{\mathscr{n}}\quad a_k\left(\mathscr{n}\right) = \frac{1-\mathscr{n}}{2 \mathscr{n}}+\frac{k-1}{\mathscr{n}} \tag{11}\label{eq:11} \end{align}
The second product in \eqref{eq:6} is done with the duplication formula Wikipedia
\begin{align} \prod _{j=1}^\mathscr{n} \Gamma\left(1+a_j-a_k\right)=\prod_{j=0}^{\mathscr{n}-1} \Gamma\left(1 + \frac{k-1}{\mathscr{n}} + \frac{j}{\mathscr{n}}\right)=(2 \pi )^{\frac{\mathscr{n}-1}{2}}\mathscr{n}^{\frac{1}{2}-(\mathscr{n}-(k-1))} \,\Gamma\left(\mathscr{n}-(k-1)\right) \tag{12}\label{eq:12} \end{align}
The simplification of the first product in \eqref{eq:6} seems to be very difficult, since the sign in the gamma function for some coefficients is negative. The same derivations have to be done similar for $\mathcal{G_\texttt{2}(\mathscr{n})}$. We find a definite sum over $\mathscr{n}$ as expected, but make no progress at this point at all.

EDIT:
In the meantime, progress is done with the sum of the two Meijer - G- Functions \eqref{eq:1}:

\begin{align} \frac{2}{(2 \pi )^{\mathscr{n}-1}}\,\tag{13}\label{eq:13} G_{\mathscr{n} + 1,\mathscr{n} + 1}^{\mathscr{n} + 1,\mathscr{n} + 1}\left( 1\left\vert \begin{array}{c} 0,\frac{ \frac{1-\mathscr{n}} {2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}}\\ 0,\frac{\frac{1-\mathscr{n}}{2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}}% \end{array}% \right. \right) = \end{align}

\begin{align*} =\, \frac{\left(1-\frac{1}{\mathscr{n}}\right)}{(2 \pi )^{\mathscr{n}-1}}\,\, % G_{\mathscr{n} + 2,\mathscr{n} + 2}^{\mathscr{n} + 2,\mathscr{n} + 1}\left( 1\left\vert \begin{array}{c} \frac{\frac{1-\mathscr{n}}{2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}},0,1\\ 0,0,\frac{\frac{1-\mathscr{n}}{2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}}% \end{array}% \right. \right)\,+ \end{align*}

\begin{align*} +\, \frac{2}{(2 \pi )^{\mathscr{n}-1}}\,\,% G_{\mathscr{n} + 2,\mathscr{n} + 2}^{\mathscr{n} + 2,\mathscr{n} + 1}\left( 1\left\vert \begin{array}{c} \frac{\frac{1-\mathscr{n}}{2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}},0,1\\ 0,0,\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 2}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n}{\mathscr{n}}% \end{array}% \right. \right) \end{align*}

This identity can be generalized for the Fox H-Function, too:

\begin{align*} 2\,H_{2,2}^{2,2}\left( 1\left\vert \begin{array}{c} \left(c_1,C_1\right) ,\left(c_2,C_2\right) \\ \left(d_1,D_1\right) ,\left(d_2,D_2\right) \end{array}\right. \right) \end{align*}

\begin{align*} =\left( 1-\frac{1}{\mathscr{n}}\right)\,H_{3,3}^{3,2}\left( 1\left\vert \begin{array}{c} \left(a_1,A_1\right) ,\left(a_2,A_2\right) ,\left(a_3,A_3\right) \\ \left(b_1,B_1\right) ,\left(b_2,B_2\right) ,\left(b_3,B_3\right) \end{array}\right. \right)\,+ \end{align*}

\begin{align*} +\,\frac{2}{\mathscr{n}}\, H_{3,3}^{3,2}\left( 1\left\vert \begin{array}{c} \left(a_1,A_1\right) ,\left(a_2,A_2\right) ,\left(a_3,A_3\right) \\ \left(b_1,B_1\right) ,\left(b_2,B_2\right) ,\left(b_3 + 1,B_3\right) \end{array}\right. \right) \end{align*}

where we used the following coefficients:
\begin{align*} \left(c_1,C_1\right) =\left(d_1,D_1\right) = \left(0,1\right),\, \left(c_2,C_2\right) =\left(d_2,D_2\right) = \left(\frac{1-\mathscr{n}}{2},\mathscr{n}\right) \end{align*}

With the new relationship \eqref{eq:13}, considering the appropriate series representation, we only have to show that:

\begin{align} \frac{\sum _ {k=1}^{\mathscr{n}+1} \left(\prod _{j=1,j\neq k}^{\mathscr{n}+1}\Gamma\left(c_k-c_j \right)\right) \prod_{j=1}^{\mathscr{n}+1}\Gamma\left(1+c_j-c_k \right)}{(2\pi )^{\mathscr{n}-1}} = \end{align}

\begin{align} \frac{\pi}{2\,\mathscr{n}} \sum _{k=0}^{\mathscr{n}-1} (-1)^k \sec \left(\frac{\pi\, k} {\mathscr{n}}\,+\, \frac{\pi }{2\,\mathscr{n}}\right)+\frac{\pi}{2} \sec \left(\frac{\pi\, \mathscr{n}}{2}\right)\, \end{align} We may express again the coefficients $c_k$ and $c_j$ as function of $\mathscr{n}$
\begin{align} c_1=0, \quad c_{j/k}=\frac{\left(\frac{1-\mathscr{n}}{2}+\left(j/k-2\right)\right)}{\mathscr{n}} \quad j,k\geq 2 \end{align}

Question:

1. How can we proof the new identity by a series representation of the two Meijer G-functions?

2. What are the appropriate series representation of each of the Meijer G-functions and how can the finite sum over the secant - function be splitted?

3. Is there a closed - form expression for the product $\prod_{j=1,j\neq k}^{\mathscr{n}+1}\Gamma\left(a_k-a_j\right)$ ?

Final Question - Final question remains after progress

The proof is done, if we show that:

\begin{align} \prod _{j=0,j\neq k}^{\mathscr{n}-1} \Gamma\left(\frac{k}{\mathscr{n}}-\frac{j}{\mathscr{n}}\right)=\frac{(-1)^{1+k} \,\mathscr{n}^{\mathscr{n}-k-\frac{3}{2}} \,\,(2 \pi )^{\frac{\mathscr{n}-1}{2}}}{\Gamma\left(\mathscr{n}-k\right)} \end{align}

Answer 1

To obtain the last identity we use the Gauss's multiplication formula: \begin{equation} \Gamma\left(nz\right)=(2\pi)^{(1-n)/2}n^{nz-(1/2)}\prod_{r=0}^{n-1}\Gamma\left(z+\frac{r}{n}\right) \end{equation} where $nz\ne 0,\pm1,\pm2\ldots$. Some modifications to the product to be evaluated are to be done. First, by changing $j=n-1-s$, we have \begin{align} P&=\prod _{j=0,j\neq k}^{n-1} \Gamma\left(\frac{k}{n}-\frac{j}{n}\right)\\ &=\prod _{s=0,s\neq n-k-1}^{n-1} \Gamma\left(\frac{k-n+1}{n}+\frac{s}{n}\right) \end{align} Now, as $k-n+1$ is a non-positive integer, we shift the arguments of 1, using the recurrence formula for the Gamma function: \begin{equation} \Gamma\left(\frac{k-n+1}{n}+\frac{s}{n}\right)=\frac{\Gamma\left(\frac{k+1}{n}+\frac{s}{n}\right)}{\frac{k-n+1}{n}+\frac{s}{n}} \end{equation} Then, \begin{align} P&=n^{n-1}\frac{\prod_{s=0,s\neq n-k-1}^{n-1}\Gamma\left(\frac{k+1}{n}+\frac{s}{n}\right)}{\prod_{s=0,s\neq n-k-1}^{n-1}(k-n+1+s)} \end{align} In the numerator, we can add the factor corresponding to $s=n-k-1$ which is $\Gamma(1)=1$ to express \begin{equation} P=n^{n-1}\frac{\prod_{s=0}^{n-1}\Gamma\left(\frac{k+1}{n}+\frac{s}{n}\right)}{\prod_{s=0,s\neq n-k-1}^{n-1}(k-n+1+s)} \end{equation} The numerator can now be evaluated with the multiplication formula with $z=(k+1)/n$: \begin{equation} \prod_{s=0,s\neq n-k-1}^{n-1}\Gamma\left(\frac{k+1}{n}+\frac{s}{n}\right)=(2\pi)^{(n-1)/2}n^{-k-1/2}\Gamma(k+1) \end{equation} The factors of the denominator $D$ are \begin{equation} -(n-k-1),-(n-k-2),\ldots,-2,-1,1,2,\ldots,k-1,k \end{equation} then, \begin{equation} D=(-1)^{n-k-1}\Gamma(n-k)\Gamma(k+1) \end{equation} Finally \begin{align} P&=n^{n-1}\frac{(2\pi)^{(n-1)/2}n^{-k-1/2}\Gamma(k+1)}{(-1)^{n-k-1}\Gamma(n-k)\Gamma(k+1)}\\ &=(-1)^{n-k+1}(2\pi)^{\frac{n-1}{2}}\frac{n^{n-k-3/2}}{\Gamma(n-k)} \end{align} Which is the expected result (up to a different sign rule. There may be a typo in the OP, as the present result seems to be numercally correct).