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Here is an approach using the residue theorem. The residues of $\newcommand{\res}{\operatorname*{Res}}f(z)=\pi\cot\pi z\csc^2\omega\pi z$ at its poles are: $$\res_{z=0}f(z)=\frac{1-\omega}{3};\quad\res_{z=n}f(z)=\csc^2\omega\pi n;\quad\res_{z=n/\omega}f(z)=-\omega\csc^2\omega\pi n\quad(n\in\mathbb{Z}_{\neq 0})$$ (easy to obtain using power series; we keep in ...

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$$n\binom{2n}n=n\cdot\dfrac{(2n!)}{n! n!}=2\binom{2n-1}{n-1}$$ Now like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots?$. comparing the expansion of $(1+x)^m$ with $$2\sum_{n=1}^\infty\binom{2n-1}{n-1} (-z)^{n-1}$$ $$mx=2\binom31(-z)^{2-1}\text{ and }\dfrac{m(m-1)}2x^... 1 By the generalized binomial theorem,$$(1+4z)^{-1/2}=\sum_{n\ge0}\frac{(-1/2)_n}{n!}4^nz^n=\sum_{n\ge0}\binom{2n}{n}(-z)^n.$$Differentiating,$$-2(1+4z)^{-3/2}=\sum_{n\ge0}\binom{2n}{n}(-1)^nnz^{n-1}=\sum_{n\ge1}\binom{2n}{n}(-1)^nnz^{n-1}.$$Multiplying by z,$$-2z(1+4z)^{-3/2}=\sum_{n\ge1}\binom{2n}{n}n(-z)^n.$$1 More generally$$\sum_{k=0}^n\sum_{i=0}^ka_{ki}b_ic_{ki}=\sum_{k=0}^n\sum_{i=0}^kb_ia_{ki}c_{ki}=\sum_{0\le i\le k\le n}b_ia_{ki}c_{ki}=\sum_{i=0}^nb_i\sum_{k=i}^na_{ki}c_{ki}.$$1 Imagine the terms in a matrix indexed by k for rows and i for columns. The right side finds the total by summing along the rows instead of down the columns. 2 Intuitively best seen by recognizing the double sum as single sum$$\sum_{(i,k)\text{ with }0\le i\le k\le n} 

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Okay I think I found a solution to my question. It's actually harder than expected. Anyone who wants to verify it is gladly welcomed. First of all, notice that the property we want to found is invariant by translation (i.e we can consider $z_i=y_i-Z$ for any $Z \in \mathbb{R}^d$). Therefore we can consider that $i=1$ and $y_1=0$ without loosing generality. ...

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Let $f(x) = \sum_{n=0}^{\infty} a_n x^n$ be a series. If it converges on an open interval $(-a,a)$, then $f$ is smooth and is equal to its Taylor series \begin{align} f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n \end{align} By uniqueness of the coefficients of the Taylor series, one has \begin{align} \forall{n} \geqslant 0,~ n!a_n = f^{(n)}(0) \end{...

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