Prove that \[ \frac{\coth\pi}{1^7}+\frac{\coth(2\pi)}{2^7}+\frac{\coth(3\pi)}{3^7}+\cdots =\frac{19\pi^7}{56700}. \]
Hint: you may use \[ \operatorname{Res}_{z=0} \frac{\cot(\pi z)\coth(\pi z)}{z^7} =-\frac{19\pi^6}{14175}. \]
Solution
1. Set\[ h(z)=\frac{\cot(\pi z)\coth(\pi z)}{z^7}. \]Besides its pole at \(0\), the poles of \(h\) are the nonzero points of \(\mathbb Z\cup i\mathbb Z\).
The poles of \(\cot(\pi z)\) are the integers, each simple with residue \(1/\pi\). The poles of \(\coth(\pi z)\) are the points \(ik\), \(k\in\mathbb Z\), again each simple with residue \(1/\pi\). Away from \(0\) these two pole sets are disjoint.
2. For every integer \(n\ge1\), the four poles \(n,-n,in,-in\) all have residue\[ \frac{\coth(\pi n)}{\pi n^7}. \]
At \(z=n\), \[ \operatorname{Res}_{z=n}h =\frac1\pi\frac{\coth(\pi n)}{n^7}. \] Since both \(\coth\) and \(z^7\) are odd, the same value is obtained at \(z=-n\).
At \(z=in\), use \[ \cot(iw)=-i\coth(w), \qquad i^7=-i. \] Then \[ \operatorname{Res}_{z=in}h =\frac1\pi\frac{\cot(\pi in)}{(in)^7} =\frac1\pi\frac{-i\coth(\pi n)}{-in^7} =\frac{\coth(\pi n)}{\pi n^7}. \] Oddness gives the same value at \(z=-in\).
3. Let \(Q_N\) be the positively oriented boundary of the square\[ |\operatorname{Re}z|\le N+\frac12, \qquad |\operatorname{Im}z|\le N+\frac12. \]Then\[ \lim_{N\to\infty}\int_{Q_N}h(z)\,dz=0. \]
The boundary of \(Q_N\) stays a fixed positive distance from both pole lattices \(\mathbb Z\) and \(i\mathbb Z\). More explicitly, for \(z=x+iy\), \[ |\cot(\pi z)|^2 =\frac{\cos^2(\pi x)+\sinh^2(\pi y)}{\sin^2(\pi x)+\sinh^2(\pi y)}, \] and \[ |\coth(\pi z)|^2 =\frac{\sinh^2(\pi x)+\cos^2(\pi y)}{\sinh^2(\pi x)+\sin^2(\pi y)}. \] On the vertical sides \(x=\pm(N+1/2)\) the first expression is at most \(1\) and the second is uniformly bounded; on the horizontal sides \(y=\pm(N+1/2)\) the second expression is at most \(1\) and the first is uniformly bounded. Thus for some constant \(C\) independent of \(N\), \[ |h(z)|\le \frac{C}{|z|^7} \le \frac{C}{(N+1/2)^7} \qquad(z\in Q_N). \] The perimeter of \(Q_N\) is \(8(N+1/2)\), so \[ \left|\int_{Q_N}h(z)\,dz\right| \le \frac{8C}{(N+1/2)^6}\longrightarrow0. \]
4. The desired series equals \(19\pi^7/56700\).
By the residue theorem and 2, \[ \frac1{2\pi i}\int_{Q_N}h(z)\,dz =\operatorname{Res}_{z=0}h +\frac4\pi\sum_{n=1}^{N}\frac{\coth(\pi n)}{n^7}. \] The series converges absolutely because \(\coth(\pi n)\) is bounded and \(\sum n^{-7}\) converges. Hence 3 and the residue supplied in the hint give \[ 0 =-\frac{19\pi^6}{14175} +\frac4\pi\sum_{n=1}^{\infty}\frac{\coth(\pi n)}{n^7}. \] Therefore \[ \sum_{n=1}^{\infty}\frac{\coth(\pi n)}{n^7} =\frac{\pi}{4}\frac{19\pi^6}{14175} =\boxed{\frac{19\pi^7}{56700}}. \]