I'm an amateur/hobbyist mathematician, and I found this interesting relationship about 6 years ago, but haven't ever quite understood it! I feel like this is related to how Digamma and Zeta are connected, but I haven't seen this directly expressed anywhere before. Could someone explain this relationship, or other ways of expressing it?
$$\sum_{k=1}^{\infty} \left( \frac{ \left( n^{k} - 1 \right)}{ \left( n + 1 \right)^{k}} \zeta \left( k+1\right) \right) = \psi \left( \frac{n}{n+1} \right) - \psi \left( \frac{1}{n+1} \right) = \pi \cot \left( \frac{\pi}{n+1} \right)$$
Also expressed as:
$$\sum_{k=1}^{\infty} \left( \frac{n^k - 1}{ \left( n+1 \right)^{k}} \sum_{i=1}^{\infty} \frac{1}{i^{k+1}} \right) = \psi \left( \frac{n}{n+1} \right) - \psi \left( \frac{1}{n+1}\right)$$
For $|z|<1$ the Digamma function has the Taylor series, $$\sum_{k\in\mathbb{N}}\zeta(k+1)z^k=-\gamma-\psi(1-z)$$ as seen here, where $\gamma$ is the Euler–Mascheroni constant.
So for the first equality,
$$\begin{align*} \sum_{k\in\mathbb{N}} \left[ \frac{ n^{k} - 1 }{ \left( n + 1 \right)^{k}} \zeta \left( k+1\right) \right] &= \sum_{k\in\mathbb{N}}\left[\left(\frac{n}{n+1}\right)^k\zeta(k+1)\right]-\sum_{k\in\mathbb{N}}\left[\frac{1}{(n+1)^k}\zeta(k+1)\right] \\ &=-\gamma-\psi\left(\frac{1}{n+1}\right)-\left(-\gamma-\psi\left(\frac{n}{n+1}\right)\right) \\ &=\psi\left(\frac{n}{n+1}\right)-\psi\left(\frac{1}{n+1}\right) \end{align*}$$ and the second equality is a direct result of the Reflection formula for the Digamma function. $$\begin{align*} \psi\left(\frac{n}{n+1}\right)-\psi\left(\frac{1}{n+1}\right)&=\psi\left(1-\frac{1}{n+1}\right)-\psi\left(\frac{1}{n+1}\right) \\ &=\pi\cot\left(\frac{\pi}{n+1}\right) \end{align*}$$
For completeness, I will also show the Reflection formula.
Recall that for $z\in\mathbb{C}$, $$\psi(z+1)=\psi(z)+\frac{1}{z}$$ this comes from taking the derivative of the relation $\Gamma(z+1)=z\Gamma(z)$, then using $\psi(z)=\Gamma'(z)/\Gamma(z)$. So we have, $$\psi(1-z)-\psi(z)=\psi(-z)-\psi(z)-\frac{1}{z}$$ whence by the series representation of the Digamma function we obtain, $$\psi(1-z)-\psi(z)=-\gamma+\sum_{k\in\mathbb{N_0}}\left(\frac{1}{k+1}-\frac{1}{k-z}\right)+\gamma-\sum_{k\in\mathbb{N_0}}\left(\frac{1}{k+1}-\frac{1}{k+z}\right)-\frac{1}{z}$$ which reduces to our desired expression. $$\begin{align*} \psi(1-z)-\psi(z)&=\sum_{k\in\mathbb{N_0}}\left(\frac{2z}{z^2-k^2}\right)-\frac{1}{z} \\ &=\sum_{k\in\mathbb{N}}\left(\frac{2z}{z^2-k^2}\right)+\frac{1}{z} \\ &=\pi\cot(\pi z) \end{align*}$$ The last equality comes from $(4.22.3)$ (Mittag-Leffler Cotangent), with $z=\pi z$ and multiplied by a factor of $\pi$.