[proofplan] We start from the classical partial fraction expansion for $\pi\cot(\pi w)$, interpreted by symmetric summation, and differentiate it $k-1$ times. Since $k \geq 4$, the differentiated lattice sum is absolutely and locally uniformly convergent away from the integers, so the differentiated symmetric expansion becomes an ordinary series. On the upper half-plane, we rewrite $\pi\cot(\pi z)$ as a normally convergent geometric [Fourier series](/page/Fourier%20Series) in $q = e^{2\pi i z}$ and differentiate termwise. Comparing the two differentiated formulae gives the stated coefficient. [/proofplan] [step:Differentiate the cotangent partial fraction expansion into an ordinary lattice sum] Define the [meromorphic function](/page/Meromorphic%20Function) \begin{align*} F: \mathbb{C} \setminus \mathbb{Z} &\to \mathbb{C} \\ w &\mapsto \pi\cot(\pi w). \end{align*} We use the classical cotangent partial fraction expansion, interpreted by symmetric summation, \begin{align*} F(w)=\lim_{N\to\infty}\sum_{n=-N}^{N}\frac{1}{w+n} \end{align*} for every $w \in \mathbb{C}\setminus\mathbb{Z}$ (citing a result not yet in the wiki: Mittag-Leffler partial fraction expansion for $\pi\cot(\pi w)$). Fix a compact set $K \subset \mathbb{C}\setminus\mathbb{Z}$. Since $K$ has positive distance from $\mathbb{Z}$ and is bounded, the series \begin{align*} \sum_{n\in\mathbb{Z}}\frac{1}{(w+n)^k} \end{align*} converges uniformly for $w \in K$ whenever $k \geq 2$, by comparison with $\sum_{|n|\geq N}|n|^{-k}$. Therefore differentiating the symmetric partial fraction expansion $k-1$ times is valid on compact subsets of $\mathbb{C}\setminus\mathbb{Z}$. For each $n \in \mathbb{Z}$, the function \begin{align*} g_n: \mathbb{C}\setminus\{-n\} &\to \mathbb{C} \\ w &\mapsto \frac{1}{w+n} \end{align*} satisfies \begin{align*} g_n^{(k-1)}(w)=(-1)^{k-1}(k-1)!\frac{1}{(w+n)^k}. \end{align*} Hence, for every $w \in \mathbb{C}\setminus\mathbb{Z}$, \begin{align*} F^{(k-1)}(w) = (-1)^{k-1}(k-1)! \sum_{n\in\mathbb{Z}}\frac{1}{(w+n)^k}. \end{align*} Equivalently, \begin{align*} \sum_{n\in\mathbb{Z}}\frac{1}{(w+n)^k} = \frac{(-1)^{k-1}}{(k-1)!}F^{(k-1)}(w). \end{align*} [guided] The partial fraction expansion for $\pi\cot(\pi w)$ is not an absolutely convergent sum at order $1$, so it must be read as the symmetric limit \begin{align*} \pi\cot(\pi w)=\lim_{N\to\infty}\sum_{n=-N}^{N}\frac{1}{w+n}, \end{align*} valid for $w \in \mathbb{C}\setminus\mathbb{Z}$ (citing a result not yet in the wiki: Mittag-Leffler partial fraction expansion for $\pi\cot(\pi w)$). After differentiating, the situation improves. Fix a compact set $K \subset \mathbb{C}\setminus\mathbb{Z}$. Because $K$ is bounded and avoids the integers, there are constants $N_0 \in \mathbb{N}$ and $C_K > 0$ such that \begin{align*} |w+n| \geq \frac{|n|}{2} \end{align*} for all $w \in K$ and all integers $n$ with $|n|\geq N_0$. Thus \begin{align*} \left|\frac{1}{(w+n)^k}\right| \leq \frac{2^k}{|n|^k} \end{align*} for $w \in K$ and $|n|\geq N_0$. Since $k \geq 4$, in particular $k>1$, the numerical series $\sum_{|n|\geq N_0}|n|^{-k}$ converges. The Weierstrass test gives [uniform convergence](/page/Uniform%20Convergence) of \begin{align*} \sum_{n\in\mathbb{Z}}\frac{1}{(w+n)^k} \end{align*} on $K$. Now define, for each $n \in \mathbb{Z}$, \begin{align*} g_n: \mathbb{C}\setminus\{-n\} &\to \mathbb{C} \\ w &\mapsto \frac{1}{w+n}. \end{align*} A direct repeated differentiation gives \begin{align*} g_n^{(k-1)}(w)=(-1)^{k-1}(k-1)!\frac{1}{(w+n)^k}. \end{align*} The locally uniform convergence just verified is exactly what is needed to differentiate the symmetric partial fraction expansion $k-1$ times away from its poles. Therefore \begin{align*} F^{(k-1)}(w) = \sum_{n\in\mathbb{Z}}g_n^{(k-1)}(w) = (-1)^{k-1}(k-1)! \sum_{n\in\mathbb{Z}}\frac{1}{(w+n)^k}. \end{align*} Solving this identity for the lattice sum gives \begin{align*} \sum_{n\in\mathbb{Z}}\frac{1}{(w+n)^k} = \frac{(-1)^{k-1}}{(k-1)!}F^{(k-1)}(w). \end{align*} [/guided] [/step] [step:Expand the cotangent as a normally convergent Fourier series on $\mathbb{H}$] Let \begin{align*} Q: \mathbb{H} &\to \mathbb{C} \\ z &\mapsto e^{2\pi i z}. \end{align*} For $z \in \mathbb{H}$, one has $|Q(z)|=e^{-2\pi\operatorname{Im}(z)}<1$. Using the identity \begin{align*} \cot(\pi z) = i\,\frac{e^{2\pi i z}+1}{e^{2\pi i z}-1}, \end{align*} we obtain \begin{align*} F(z) = \pi i\,\frac{Q(z)+1}{Q(z)-1} = -\pi i\,\frac{1+Q(z)}{1-Q(z)}. \end{align*} Since $|Q(z)|<1$, \begin{align*} \frac{1+Q(z)}{1-Q(z)} = 1+2\sum_{r=1}^{\infty}Q(z)^r. \end{align*} Thus \begin{align*} F(z) = -\pi i-2\pi i\sum_{r=1}^{\infty}e^{2\pi i r z}. \end{align*} Moreover, if $K \subset \mathbb{H}$ is compact, then there is a number $\rho_K \in (0,1)$ such that $|Q(z)|\leq \rho_K$ for every $z \in K$. For every integer $m \geq 0$, \begin{align*} \left|(2\pi i r)^m e^{2\pi i r z}\right| \leq (2\pi r)^m \rho_K^r \end{align*} for every $z \in K$ and every $r \geq 1$, and the numerical series $\sum_{r=1}^{\infty}(2\pi r)^m\rho_K^r$ converges. Hence the Fourier series for $F$ and all of its termwise derivatives converge uniformly on $K$. [guided] The upper half-plane is the region where the exponential variable has modulus strictly less than $1$. Define \begin{align*} Q: \mathbb{H} &\to \mathbb{C} \\ z &\mapsto e^{2\pi i z}. \end{align*} If $z = x+iy$ with $x,y \in \mathbb{R}$ and $y>0$, then \begin{align*} |Q(z)|=|e^{2\pi i x}e^{-2\pi y}|=e^{-2\pi y}<1. \end{align*} This is why the geometric expansion below is valid precisely on $\mathbb{H}$. Using the exponential formula for cotangent, \begin{align*} \cot(\pi z) = i\,\frac{e^{2\pi i z}+1}{e^{2\pi i z}-1}, \end{align*} we rewrite $F(z)=\pi\cot(\pi z)$ as \begin{align*} F(z) = \pi i\,\frac{Q(z)+1}{Q(z)-1} = -\pi i\,\frac{1+Q(z)}{1-Q(z)}. \end{align*} Since $|Q(z)|<1$, the geometric series gives \begin{align*} \frac{1}{1-Q(z)} = \sum_{r=0}^{\infty}Q(z)^r. \end{align*} Multiplying by $1+Q(z)$ yields \begin{align*} \frac{1+Q(z)}{1-Q(z)} = 1+2\sum_{r=1}^{\infty}Q(z)^r. \end{align*} Therefore \begin{align*} F(z) = -\pi i-2\pi i\sum_{r=1}^{\infty}e^{2\pi i r z}. \end{align*} We also need termwise differentiation. Let $K \subset \mathbb{H}$ be compact. Since the continuous function $z \mapsto |Q(z)|$ is strictly less than $1$ on $K$, there exists $\rho_K \in (0,1)$ such that $|Q(z)|\leq \rho_K$ for all $z \in K$. For an integer $m \geq 0$, the $m$-th derivative of $e^{2\pi i r z}$ is $(2\pi i r)^m e^{2\pi i r z}$, and therefore \begin{align*} \left|(2\pi i r)^m e^{2\pi i r z}\right| \leq (2\pi r)^m \rho_K^r. \end{align*} The series $\sum_{r=1}^{\infty}(2\pi r)^m\rho_K^r$ converges because exponential decay dominates every polynomial power of $r$. The Weierstrass test gives uniform convergence on $K$ for the Fourier series and for each differentiated series. Hence differentiating termwise on compact subsets of $\mathbb{H}$ is justified. [/guided] [/step] [step:Differentiate the Fourier series and compare the two formulae] Taking $m=k-1$ in the termwise differentiation justified above, the constant term $-\pi i$ differentiates to $0$, and \begin{align*} F^{(k-1)}(z) = -2\pi i \sum_{r=1}^{\infty}(2\pi i r)^{k-1}e^{2\pi i r z} = -(2\pi i)^k \sum_{r=1}^{\infty}r^{k-1}e^{2\pi i r z}. \end{align*} Substituting this into the differentiated partial fraction identity at $w=z$ gives \begin{align*} \sum_{n\in\mathbb{Z}}\frac{1}{(z+n)^k} &= \frac{(-1)^{k-1}}{(k-1)!}F^{(k-1)}(z) \\ &= \frac{(-1)^{k-1}}{(k-1)!} \left( -(2\pi i)^k \sum_{r=1}^{\infty}r^{k-1}e^{2\pi i r z} \right) \\ &= \frac{(-1)^k(2\pi i)^k}{(k-1)!} \sum_{r=1}^{\infty}r^{k-1}e^{2\pi i r z} \\ &= \frac{(-2\pi i)^k}{(k-1)!} \sum_{r=1}^{\infty}r^{k-1}e^{2\pi i r z}. \end{align*} This is the desired identity. The absolute convergence of the left-hand series follows from comparison with $\sum_{|n|\geq 1}|n|^{-k}$, and the absolute convergence of the right-hand series follows from $|e^{2\pi i r z}|=e^{-2\pi r\operatorname{Im}(z)}$ together with exponential decay. [/step]