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Let $k$, $d$ and $p$ be strictly positive integers such that $k \ge 2 d$. Define a following multiple sum: \begin{equation} {\mathfrak S}^{(d)}(k,p) := \sum\limits_{\begin{array}{rrr} 1 \le & i_d & \le k-2 d+1 \\ 2+i_d \le & i_{d-1} & \le k-2 d+3 \\ 2+i_{d-1} \le & i_{d-2} & \le k-2 d+5\\ \vdots & \vdots & \vdots \\ 2+i_3 \le & i_2 & \le k-3 \\ 2+ i_2 \le & i_1 & \le k-1\end{array}} \prod\limits_{q=1}^d \frac{1}{\binom{2 i_q+2p-1}{2 p} } \end{equation} This multiple sum appears in the context of a certain recurrence relation with time dependent coefficients(see Closed form of an inhomogeneous non-constant recurrence relation). We have calculated the result for $d=1$. It reads: \begin{eqnarray} &&{\mathfrak S}^{(1)}(k,p) =(2p)\cdot \left.\left[\sum\limits_{l=0}^{2p-2} \binom{2p-2}{l} (-1)^l \cdot \Phi(-1,1,1+n+l)\right] \right.^{n=1}_{n=2k-1} \end{eqnarray} Here $\Phi$ is the Lerch transcendent and $\left. ._n \right|^{n=a}_{n=b} := ._a - ._b$.

Now the obvious question would be what is the result for arbitrary values of $d$.

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The formal solution is given below: \begin{eqnarray} {\mathfrak S}^{(d)}_{(k,p)} = (2p)^d \sum\limits_{q=0}^d (-1)^q \int\limits_{[0,1]^d} \frac{\left(\prod\limits_{\xi=0}^{d-1-q} t_\xi^{4 \xi+1}\right) \cdot \left(\prod\limits_{\xi=d-q}^{d-1} t^{2(k-d+\xi)+1} \right) \cdot\prod\limits_{\xi=0}^{d-1} (1-t_\xi)^{2 p-1}}{\prod\limits_{\xi=0}^{d-1-q} (1- \prod\limits_{\eta=\xi}^{d-1-q} t^2_\eta)\cdot \prod\limits_{\xi=d-q}^{d-1}(1- \prod\limits_{\eta=d-q}^{\xi} t^2_\eta)} \prod\limits_{\xi=0}^{d-1} d t_\xi \end{eqnarray} If $d$ is big the result above is of limited use for numerical calculations since the multidimensional integrals are hard to compute . However the result is quite useful for analyzing the limit $k\rightarrow \infty$ . We have: \begin{equation} {\mathfrak S}^{(d)}_{p}:=\lim\limits_{k\rightarrow \infty} {\mathfrak S}^{(d)}_{(k,p)} = (2p)^d \int\limits_{[0,1]^d} \prod\limits_{\xi=0}^{d-1} \frac{t_\xi^{4 \xi+1} \cdot (1-t_\xi)^{2p-1}}{\left(1- \prod\limits_{\eta=\xi}^{d-1} t_\eta^2\right)} d\vec{t} \end{equation} Now by expanding the denominator in an infinite series , carrying out the integration term by term and finally by resumming the resulting series we can actually obtain quite neat expressions for the limits in question. We have: \begin{eqnarray} {\mathfrak S}^{(d)}_p = \left\{ \begin{array}{rrr} (2p) \sum\limits_{l=0}^{2p-2} \binom{2p-2}{l} \left[\sum\limits_{l_1=1}^{l+1} \frac{(-1)^{l_1-1}}{l_1}\right] - p \cdot 2^{p-1} \cdot \log(2) & \mbox{if $d=1$} \\ \cdots & \mbox{if $d=2$} \end{array} \right. \end{eqnarray} We will finish those calculations at a later stage.

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