I wanted to solve the zeta function for an undefined period "$d$". So for every $d\ge2$.

$$\zeta(-s)= \frac{1}{(d^{s+1}-1)}\sum_{m=1}^{\infty} \frac{1}{2^{m+1}}\sum^{m}_{j=1} (-1)^{j+1}(j)^s\dbinom{m}{j} \sum_{k=0}^{m+1}\dbinom{m+1}{k} \sum^{d-1}_{p=1} \frac{e^{\frac{2i\pi p}{d}}*i^k}{\tan^k(\frac{\pi p}{d})} $$

So far so good, this works fine. Now I want to solve the sum because I believe that's how I make progress.

$$\sum^{d-1}_{p=1}\frac{e^{\frac{2i\pi p}{d}}*i^k}{\tan^k(\frac{\pi p}{d})} $$ I already asked a question about the summation, Geometry formulas, how to show identities.

A few try's where I left out the sign, $(-1)^{\frac{k^2+k}{2}}$:

$$ \begin{align} k(0)&=0-1 \\ k(1)&=(d-2)\\ k(2)&=\frac{(d-2)(d-4)}{3}\\ k(3)&=\frac{(d-2)(2d-5)}{3}\\ k(4)&=\frac{(d-2)(d-4)(d^2+6d-22)}{45}\\ k(5)&=\frac{(d-2)(2d^3+4d^2-62d+101)}{45} \end{align} $$

Question, what's the general function?

Next question will be, but I'd like to give that my own shot, how to I solve the next summation.

An example for $s=1$: $$\frac{(k(0)-2k(1)-k(2))}{4(d^2-1)} = -\frac{1}{12}$$

PS: I know you can just fill in $d=2$ and get the "normal" zeta function, but that's the goal is to extend it so i want to find an expersioin for every d.


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