Do we know the limit of $\sum\limits_{k=1}^{n}\frac{1}{(a i k+b)^2}$? I am not sure if the closed- form of this limit is known ($i$ is the imaginary one):
$$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{(a i k+b)^2}$$
If it's not, then yay, I found it. But I think it's very unlikely that it's not already known, as with anything else I've found out.
Edit:
All right, I just checked the answer with the Polygamma on Mathematica, and it doesn't give it a closed-form, but I do have and am adding it here.
Please don't vandalize the rating of this post otherwise I won't be able to add the details.

$$\Re\left(\lim_{n\to\infty}\sum_{k=1}^n\frac1{(aik+b)^2}\right)=-\frac1{2b^2}+\frac1{2(1-e^{-2\pi b/a})^2}\left(\frac{2\pi}a\right)^2e^{-2\pi b/a}$$

Edit:
Attending to the request of the mods, I want to clarify that the question is about if a closed-form for the below was already known (see my answer for the real part):
$$\lim_{n\to\infty}\sum_{j=1}^{n}\frac{1}{(a i j k+b)^k}$$
 A: So we have 
$$\frac1{(aik+b)^2}=\frac{1}{b^2-a^2k^2+2abki}$$
$$=\frac{b^2-a^2k^2-2abki}{(b^2-a^2k^2)^2+(2abk)^2}$$
$$=\frac{b^2-a^2k^2-2abki}{b^4-2a^2b^2k^2+a^4k^4+4a^2b^2k^2}$$
$$=\frac{b^2-a^2k^2-2abki}{b^4+2a^2b^2k^2+a^4k^4}$$
$$=\frac{b^2-a^2k^2-2abki}{(b^2+a^2k^2)^2}$$
$$=\frac{b^2-a^2k^2}{(b^2+a^2k^2)^2}-\frac{2abki}{(b^2+a^2k^2)^2}$$
$$\therefore \Re\left(\lim_{n\to\infty}\sum_{k=1}^n\frac1{(aik+b)^2}\right)=\lim_{n\to\infty}\sum_{k=1}^n \frac{b^2-a^2k^2}{(b^2+a^2k^2)^2}$$
$$=\lim_{n\to\infty}\sum_{k=1}^n \frac{d}{da}\Bigg(\frac{a}{b^2+a^2k^2}\Bigg)$$
$$=\frac{d}{da}\Bigg(\lim_{n\to\infty}\sum_{k=1}^n \frac{a}{b^2+a^2k^2}\Bigg)$$
$$=\frac{d}{da}\Bigg(\lim_{n\to\infty}\frac{1}{a}\sum_{k=1}^n \frac{1}{(b/a)^2+k^2}\Bigg)$$
$$=\frac{d}{da}\Bigg(\frac{1}{a}\frac{\pi (\frac{b}{a})\coth{(\frac{\pi b}{a})} - 1}{2 (\frac{b}{a})^2}\Bigg)$$
$$=\frac{d}{da}\Bigg(\frac{\pi b \coth{(\frac{\pi b}{a})} - a}{2 b^2}\Bigg)$$
$$=\frac{\pi^2}{2a^2\sinh^2{(\frac{\pi b}{a})}}-\frac{1}{2b^2}$$
Which is the same as your result with some rearrangement.
A: Below is the formula I found for the general case. If someone knows if this result is already known, please let me know by posting links to papers or pages. 
I will claim it as a different proof nonetheless on my arXiv paper, which I will post here when I'm done:
\begin{multline}\nonumber
h(x)=\sum _{k=2}^{\infty}\sum _{j=1}^{\infty}\frac{x^k}{(a j+b)^k}=-\frac{x^2}{2b(b-x)}+\frac{\pi x}{2a}\csc{\frac{\pi b}{a}}\csc{\frac{\pi (b-x)}{a}}\sin{\frac{\pi x}{a}}\\-\frac{x \pi}{a}\int _0^1\left(\csc{\frac{2 \pi (b-x)}{a}}\sin{\frac{2 \pi (b-x)u}{a}}-\csc{\frac{2
\pi b}{a}}\sin{\frac{2 \pi  b u}{a}}\right)\cot{\pi u}\,du
\end{multline}
On the limits of a generalized harmonic progression:
https://arxiv.org/abs/1902.06885
