## On Riemann-like zeta functions

Given the Riemann zeta function $\zeta(s)$, there is the nice equality,

\begin{aligned}&\sum_{m=1}^\infty\big[\zeta(2m)-1] = \frac{3}{4}\end{aligned}

It can be shown that,

\begin{aligned}&\sum_{m=1}^\infty\big[\zeta(pm)-1] = \sum_{k=2}^\infty \frac{1}{k^p-1}\end{aligned}

Consider the following evaluations,

\begin{aligned}&\sum_{k=2}^\infty \frac{1}{k^2-1} = \frac{3}{4} = 0.75\\ &\sum_{k=2}^\infty \frac{1}{k^2+1} = -1+\frac{\pi\text{coth}(\pi)}{2} = 0.5766\dots\end{aligned}

In general, given a root of unity, $\omega_p = e^{2\pi i/p}$, then,

\begin{aligned}&\sum_{k=2}^\infty \frac{1}{k^p-a} = -\frac{a^{1/p}}{ap}\sum_{j=1}^p \omega_p^j\, \psi(2-a^{1/p} \omega_p^j)\end{aligned}

for integer p > 1, any non-zero real or complex a, and where $\psi(z)$ is the digamma function. Thus, since roots of unity are involved, the formula uses complex terms even though, as the two examples show, the sum may be real.  But it turns out for real a and even powers p, it can be expressed using only real terms.  First,

\begin{aligned}&\sum_{k=2}^\infty \frac{1}{k^p-a} = \frac{1-3a}{2a(1-a)} -\frac{a^{1/p}\,\pi}{ap}\sum_{j=1}^{p/2} \omega_p^j\, \cot(\pi a^{1/p} \omega_p^j)\end{aligned}

for even p and any non-zero a except a = 1,  which is given by the special case,

\begin{aligned}&\sum_{k=2}^\infty\frac{1}{k^p-1} = \frac{2p-1}{2p}-\frac{\pi}{p}\sum_{j=1}^{p/2-1}\omega_p^j\,\cot(\pi\omega_p^j)\end{aligned}

But one can split the cotangent function into its real and imaginary parts as,

\begin{aligned}&\cot(\pi u\, e^{2\pi i n}) = \frac{-\sin(2\pi u\cos(2\pi n))+i \text{sinh}(2\pi u\sin(2\pi n)) }{\cos(2\pi u\cos(2\pi n))-\text{cosh}(2\pi u \sin(2\pi n))}\end{aligned}

hence cancel out the conjugate terms and leave only the real parts.  For example, we have,

\begin{aligned}&\sum_{k=2}^\infty \frac{1}{k^4-1} = \frac{1}{8}\big(7-2\pi\text{coth}(\pi)\big) = 0.0866\dots\\ &\sum_{k=2}^\infty \frac{1}{k^6-1} = \frac{1}{12}\big(11-2\pi\sqrt{3}\text{tanh}(\tfrac{\pi\sqrt{3}}{2})\big)= 0.0175\dots\end{aligned}

and so on. It is reminiscent of the situation with the zeta function,

\begin{aligned}&\sum_{k=1}^\infty \frac{1}{k^p} = \zeta(p)\end{aligned}

which has a closed-form solution only for even p, and is expressed by the real $\pi^p$  and Bernoulli numbers.  It makes me wonder if there is  a closed-form formula for $\zeta(p)$  involving the roots of unity.