Let $a>0$ and $x\in(-1,1)$. Can we give a closed form expression for $\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}$? Note that the series is convergent. There should be an expression in terms of the cotangent.
My goal is to find a (sharp) upper bound $c(x)$ of the series.
On
Observe \begin{align} \sum^\infty_{k=-\infty}\frac{1}{a+(x+k)^2} =&\ \frac{1}{a+(1+x)^2}+\frac{1}{a+x^2}+\frac{1}{a+(1-x)^2}\\ &+\sum^\infty_{k=1} \frac{1}{a+(k+(1+x))^2}+\sum^{\infty}_{k=1} \frac{1}{a+(k+(1-x))^2}\\ \leq&\ \frac{1}{a+(1+x)^2}+\frac{1}{a+x^2}+\frac{1}{a+(1-x)^2}\\ &\ +\int^\infty_0\frac{dk}{a+(k+(1+x))^2}+ \int^\infty_{0} \frac{dk}{a+(k+(1-x))^2}\\ \leq&\ \frac{1}{a+(1+x)^2}+\frac{1}{a+x^2}+\frac{1}{a+(1-x)^2}\\ &\ +\frac{1}{\sqrt{a}}\left(\tan^{-1}\left( \frac{\sqrt{a}}{1+x}\right)+\tan^{-1}\left( \frac{\sqrt{a}}{1-x}\right)\right) \end{align}
Assuming that you enjoy special functions $$\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}=\frac{-\psi \left(-x-\sqrt{-a}+1\right)+\psi \left(-x+\sqrt{-a}+1\right)-\psi \left(x-\sqrt{-a}\right)+\psi \left(x+\sqrt{-a}\right)}{2 \sqrt{-a}}$$ wich can simplify as $$\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}=-\frac{\pi \left(\cot \left(\pi \left(\sqrt{-a}-x\right)\right)+\cot \left(\pi \left(\sqrt{-a}+x\right)\right)\right)}{2 \sqrt{-a}}$$ and, since $a >0$ $$\color{blue}{\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}=\frac{\pi \left(\coth \left(\pi \left(\sqrt{a}-i x\right)\right)+\coth \left(\pi \left(\sqrt{a}+i x\right)\right)\right)}{2 \sqrt{a}}}$$ what you can simplify using $$\coth(A+B)+\coth(A-B)=\frac{2 \sin (2 A)}{\cos (2 B)-\cos (2 A)}$$
Edit
In order to keep the results in the answer, I reproduce here what you wrote in comments.
So, we have finally $$\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}=\frac \pi {\sqrt a}\,\,\frac{\sinh \left(2 \pi \sqrt{a}\right)}{\cosh \left(2 \pi \sqrt{a}\right)-\cos(2 \pi x)}$$ but I think that you cannot make at the same time $a \to 0$ and $x\to \pm 1$ without avoiding $\infty$. In the post, remember that you did precise $x\in(-1,1)$ and not $x\in[-1,1]$.