EDIT:
The question here is not for the reader to laboriously scan all the working, but rather to suggest ways to continue the train of thought; for example, there is perhaps a closed form of the cotangent sum.
OP:
As introductory examples, consider, where $\psi$ denotes digamma, $\gamma$ denotes Euler's constant, and the multiplication theorem of Gauss is recalled:
$$\begin{align}\psi(n+1/2)&=2\cdot\psi(2n+1)-\psi(n+1)-2\ln2\\&=2(-\gamma+\sum_{k=1}^{2n}\frac{1}{k})-(-\gamma+\sum_{k=1}^n\frac{1}{k})-2\ln2\\&=-\gamma+2\left(-\ln2+\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{2k}\right)\\&=-\gamma+2\left(-\ln2+\sum_{k=1}^{n}\frac{1}{2k-1}\right)\\\sum_{k=1}^\infty\frac{1}{k}-\frac{1}{k+n-1/2}&=-2\ln2+\sum_{k=1}^n\frac{2}{2k-1}\\\sum_{k=1}^\infty\frac{1}{k}-\sum_{k=n}^\infty\frac{2}{2k-1}&=-2\ln2+\sum_{k=1}^n\frac{2}{2k-1}\\\sum_{k=1}^\infty\frac{1}{k}-\frac{2}{2k-1}&=-2\ln2\end{align}$$
This result can be reached using more trivial means however. Let us consider instead digamma of thirds, along with the reflection formula and the harmonic recurrence $\psi(s+1)=\frac{1}{s}+\psi(s)$:
$$\begin{align}\psi(n+1/3)+\psi(n+2/3)&=3\cdot\psi(3n+1)-\psi(n+1)-3\ln3\\\psi(1/3)+\psi(2/3)+\sum_{k=1}^n\left(\frac{3}{3k-1}+\frac{3}{3k-2}\right)&=-2\gamma-3\ln3+3\left(\sum_{k=1}^{3n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{3k}\right)\\2\cdot\psi(1/3)+\frac{1}{\sqrt{3}}\pi&=-2\gamma-3\ln3+0\\2\cdot\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{3}{3k-2}\right)&=-3\ln3-\frac{1}{\sqrt{3}}\pi\\\sum_{k=1}^\infty\frac{1}{k}-\frac{3}{3k-2}&=-\frac{1}{2}\left(3\ln3+\frac{1}{\sqrt{3}}\pi\right)\end{align}$$
This is now more striking. I could not see any way to simplify the left summand using any telescoping tricks, but I did try to press on with the same pattern.
EDIT: The formatting broke in the transfer from question-writing to posting! I regretfully do not have the Tex skills to make this look any nicer. An edit or suggestion is welcome.
First consider even integers; I try $m=2d$ for $d\gt1$. We have: $$\small{\begin{align}\sum_{i=1}^m\psi(n+i/m)&=m\cdot\psi(mn+1)-m\ln m\\\psi(n+1/2)+\sum_{i=1}^{d-1}(\psi(n+i/m)+\psi(n+1-i/m))&=(1-m)\gamma-m\ln m+m\left(\sum_{k=1}^{mn}\frac{1}{k}-\sum_{k=1}^n\frac{1}{mk}\right)\\-\gamma-2\ln2+\sum_{k=1}^n\frac{2}{2k-1}+\sum_{i=1}^{d-1}\left(\psi(i/m)+\psi(1-i/m)+\sum_{k=1}^n\left(\frac{m}{km-i}+\frac{m}{km+i-m}\right)\right)&=(1-m)\gamma+m\left(\sum_{k=1}^{mn}\frac{1}{k}-\sum_{k=1}^n\frac{1}{mk}\right)\\-\gamma-2\ln2+\sum_{k=1}^n\frac{2}{2k-1}+\sum_{k=1}^{d-1}\left(2\cdot\psi(i/m)+\pi\cot\pi\cdot\frac{i}{m}\right)+\sum_{i=1}^{d-1}\sum_{k=1}^n\left(\frac{m}{mk-i}+\frac{m}{mk+i-m}\right)&=(1-m)\gamma-m\ln m+m\left(\sum_{k=1}^{mn}\frac{1}{k}-\sum_{k=1}^n\frac{1}{mk}\right)\\(2-m)\gamma+2\ln2-m\ln m+m\left(\sum_{k=1}^{mn}\frac{1}{k}-\sum_{k=1}^n\frac{1}{mk}\right)-m\left(\sum_{k=1}^{mn}\frac{1}{k}-\sum_{k=1}^n\left(\frac{2}{2km-m}+\frac{1}{km}\right)\right)&=\sum_{i=1}^{d-1}2\cdot\psi(i/m)+\pi\cot\pi\cdot\frac{i}{m}+\sum_{k=1}^n\frac{2}{2k-1}\\2\ln2-m\ln m+\sum_{k=1}^n\frac{2}{2k-1}&=\sum_{k=1}^{d-1}\left(\pi\cot\pi\cdot\frac{i}{m}+2\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{m}{km+i-m}\right)\right)+\sum_{k=1}^n\frac{2}{2k-1}\\2\ln2-m\ln m-\sum_{i=1}^{d-1}\pi\cot\pi\cdot\frac{i}{m}&=2\sum_{i=1}^{d-1}\sum_{k=1}^\infty\frac{1}{k}-\frac{m}{km+i-m}\end{align}}$$ Now consider odd integers; $m=2d+1$. We have:$$\small{\begin{align}\sum_{i=1}^m\psi(n+i/m)&=m\cdot\psi(mn+1)-m\ln m\\\sum_{i=1}^d(\psi(n+i/m)+\psi(n+1-i/m))&=(1-m)\gamma-m\ln m+m\left(\sum_{k=1}^{mn}\frac{1}{k}-\sum_{k=1}^n\frac{1}{mk}\right)\\\sum_{i=1}^d\left(\psi(i/m)+\psi(1-i/m)+\sum_{k=1}^n\left(\frac{m}{km-i}+\frac{m}{km+i-m}\right)\right)&=(1-m)\gamma+m\left(\sum_{k=1}^{mn}\frac{1}{k}-\sum_{k=1}^n\frac{1}{mk}\right)\\\sum_{k=1}^d\left(2\cdot\psi(i/m)+\pi\cot\pi\cdot\frac{i}{m}\right)+\sum_{i=1}^d\sum_{k=1}^n\left(\frac{m}{mk-i}+\frac{m}{mk+i-m}\right)&=(1-m)\gamma-m\ln m+m\left(\sum_{k=1}^{mn}\frac{1}{k}-\sum_{k=1}^n\frac{1}{mk}\right)\\(1-m)\gamma-m\ln m+m\left(\sum_{k=1}^{mn}\frac{1}{k}-\sum_{k=1}^n\frac{1}{mk}\right)-m\left(\sum_{k=1}^{mn}\frac{1}{k}-\sum_{k=1}^n\frac{1}{km}\right)&=\sum_{i=1}^d2\cdot\psi(i/m)+\pi\cot\pi\cdot\frac{i}{m}\\-m\ln m&=\sum_{k=1}^d\left(\pi\cot\pi\cdot\frac{i}{m}+2\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{m}{km+i-m}\right)\right)\\-m\ln m-\sum_{i=1}^d\pi\cot\pi\cdot\frac{i}{m}&=2\sum_{i=1}^d\sum_{k=1}^\infty\frac{1}{k}-\frac{m}{km+i-m}\end{align}}$$
After that messy derivation, we arrive at the even/odd results (some sort of half-baked multiplication theorem) for $d(\in\Bbb N)\gt1$:
$$\begin{align}\ln2-d\ln2d-\frac{\pi}{2}\sum_{i=1}^{d-1}\cot\left(\pi\cdot\frac{i}{2d}\right)&=\sum_{i=1}^{d-1}\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{2d}{2d(k-1)+i}\right)\\&=(d-1)\gamma+\sum_{i=1}^{d-1}\psi\left(\frac{i}{2d}\right)\\-\left(d+\frac{1}{2}\right)\ln(2d+1)-\frac{\pi}{2}\sum_{i=1}^d\cot\left(\pi\cdot\frac{i}{2d+1}\right)&=\sum_{i=1}^d\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{2d+1}{(2d+1)(k-1)+i}\right)\\&=d\gamma+\sum_{i=1}^{d}\psi\left(\frac{i}{2d+1}\right)\end{align}$$
Which amazingly is actually correct, if one plots the two sides on Desmos, but I would love to simplify it further. I have completely failed in this regard! It would be great if someone here could help me simplify any of the results I find here, and/or suggest other interesting polygamma tricks. I am proud to have found this relation, and I hope it can lead to some of the interesting series identities shown on the Wikipedia page; or perhaps it's a "new" identity not previously mentioned. I just don't know how to keep this going!
Many thanks in advance.