Do we know the limit of $\sum\limits_{k=1}^{n}\frac{1}{(a i k+b)^2}$?

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.

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