I am trying to find the pattern for the coefficients of the closed forms for this series:
$$\sum_{n=1}^\infty \left( \frac{1}{(k+n)^2+n^2}-\frac{1}{(k+1-n)^2+n^2} \right)$$
The series seems to converge for every natural $k$, and it is easy to evaluate each of the two series using this, ending up with a bunch of digamma functions, as shown here. However, when plugging the series in wolfram for some values of $k$ this is what I got:
\begin{align}
k=1: &&& \frac{-5+2\pi\tanh\left(\frac{\pi}{2}\right)-\pi\coth(\pi)}{4} \\
k=2: &&& \frac{-42-10\pi\tanh\left(\frac{3\pi}{2}\right)+15\pi\coth(\pi)}{60} \\
k=3: &&& \frac{-341+120\pi\tanh\left(\frac{3\pi}{2}\right)-90\pi\coth(2\pi)}{720} \\
k=4: &&& \frac{-3189-884\pi\tanh\left(\frac{5\pi}{2}\right)+1105\pi\coth(2\pi)}{8840} \\
k=5: &&& \frac{-58076+19890\pi\tanh\left(\frac{5\pi} {2}\right)-16575\pi\coth(3\pi)}{198900} \\
k=6: &&& \frac{-21576583-6277050\pi\tanh\left(\frac{7\pi}{2}\right)+7323225\pi\coth(3\pi)}{87878700}
\end{align}
The arguments of $\tanh$ and $\coth$ can be guessed, respectively, to $\frac{k+(k+1\mod2)}{2}\pi$ and $\frac{k+(k\mod2)}{2}\pi$, but the other coefficients are less easy to confront. The linear sequence in the numerators ($5,42,341,\dots$) does not appear even in the OEIS, and same goes for the denominators.
However, it is possible to guess the coefficients of $\tanh$ and $\coth$. Hence the question:
How to show that
\begin{align}
\sum_{n=1}^\infty \left( \frac{1}{(k+n)^2+n^2}-\frac{1}{(k+1-n)^2+n^2} \right)= \\
=-\frac{a_k}{b_k}+\frac{\pi}{2k}\tanh\left(\frac{k}{2}\pi\right)-\frac{\pi}{2k+2}\coth\left(\frac{k+1}{2}\pi\right) \text{,} &&& k\ \text{odd}\\
=-\frac{a_k}{b_k}-\frac{\pi}{2k-2}\tanh\left(\frac{k+1}{2}\pi\right)+\frac{\pi}{2k}\coth\left(\frac{k}{2}\pi\right) \text{,} &&& k\ \text{even}\\
\end{align}
and what are $a_k$ and $b_k$?
I know that, through residues, one can show that $$\frac{1}{k^2}+2\sum_{n=1}^{\infty}\frac{(-1)^n}{k^2+n^2}=\frac{\pi}{k\sinh(k\pi)}$$
and I believe something similar is going on here, though I am not able to see how.
Here are the most promising attempt so far:
Appearently the result in terms of digamma functions can be simplified a lot to
$$\sum_{n=1}^\infty \left( \frac{1}{(k+n)^2+n^2}-\frac{1}{(k+1-n)^2+n^2} \right)=$$
$$=\frac{i}{2k}\left(\psi\left(1+\bar{z}k\right)-\psi\left(1+zk\right)\right)+\frac{i}{2(k+1)}\left(\psi(z-\bar{z}k)-\psi(\bar{z}-zk)\right) \tag{*}$$
where $z=\frac{1+i}{2}$
I feel like it is almost done, but I can't perform the last steps to get rid of the digamma functions.
However, I strongly believe it has to do with the following two identities for $u\in \mathbb{C}$:
$$\psi(1-u)-\psi(u)=\pi\cot(\pi u) $$
$$\psi\left(\frac12+u\right)-\psi\left(\frac12-u\right)=\pi\tan(\pi u)$$
Any ideas on how to finish?
EDIT
I removed the other attempts since they were irrelevant.
Best Answer
I finally did it!
What follows is the process that leads to the final result in terms of hyperbolic functions. Plus, two additional very interesting side-results.
I will start from the formula in $(\text{*})$: for $z=\frac{1+i}{2}$ the following holds $$\sum_{n=1}^\infty \left( \frac{1}{(k+n)^2+n^2}-\frac{1}{(k+1-n)^2+n^2} \right)=$$
$$=\frac{i}{2k}\left(\psi\left(1+\bar{z}k\right)-\psi\left(1+zk\right)\right)+\frac{i}{2(k+1)}\left(\psi(z-\bar{z}k)-\psi(\bar{z}-zk)\right)$$
and then work my way through the cancelation of the digamma functions.
Tools:
The main tools we are going to use are the following: $$\psi(1+u)=\psi(u)+\frac{1}{u} \tag{1}$$ $$\psi(1-u)-\psi(u)=\pi\cot(\pi u) \tag{2} $$ $$\psi(u+k)=\psi(u)+\sum_{n=0}^{k-1}\frac{1}{u+n}\tag{3}$$ $$\psi(-u)=\psi(u)+\frac1u+\pi\cot(\pi u)\tag{4}$$ Now, for reference purpose, let's name $$\color{red}{S_1}=\frac{i}{2k}\left(\psi\left(1+\bar{z}k\right)-\psi\left(1+zk\right)\right) \hspace{1cm} \color{blue}{S_2}=\frac{i}{2(k+1)}\left(\psi(z-\bar{z}k)-\psi(\bar{z}-zk)\right)$$ and let's start with $S_1$.
PART $1$: $S_1$
It all starts by noticing that since $z=\frac{1+i}{2}$, we have that $\bar{z}=1-z$, therefore: \begin{align} \psi(1+\bar{z}k)-\psi(1+zk)&=\psi(1+k-zk)-\psi(1+zk) \\ &\stackrel{\text{(1)}}{=}\psi(1+k-zk)-\psi(zk)-\frac{1}{zk} \\ &\stackrel{\text{(3)}}{=}\psi(1-zk)+\sum_{n=0}^{k-1}\frac{1}{1-zk+n}-\psi(zk)-\frac{1}{zk} \\ &\stackrel{\text{(2)}}{=}\pi\cot(\pi zk)+\sum_{n=0}^{k-1}\frac{1}{1-zk+n}-\frac{1}{zk} \\ \end{align} Next, plug $z$ in and simplify: $$\frac{1}{zk}=\frac1k-\frac{i}{k}$$ $$\frac{1}{n+1-zk}=\frac{2}{2n+2-(1+i)k}=\frac{4n+4-2k}{(2n+2-k)^2+k^2}+i\frac{2k}{(2n+2-k)^2+k^2}$$ $$\cot(\pi zk)=\cot\left(\frac{\pi}{2}(1+i)k\right)=-\frac{\sin(k\pi)}{\cos(k\pi)-\cosh(k\pi)}+i\frac{\sinh(k\pi)}{\cos(k\pi)-\cosh(k\pi)}$$ and finally, putting these values back in $S_1$, together with the fact that $\sin(k\pi)=0$ and $\cos(k\pi)=(-1)^k$, we get \begin{align} \color{red}{S_1}&=\frac{i}{2k}(\psi(1+\bar{z}k)-\psi(1+zk)) \\ &=-\frac{\pi}{2k}\frac{\sinh(k\pi)}{(-1)^k-\cosh(k\pi)}-\frac{1+i}{2k^2}+\frac{i}{k}\sum_{n=0}^{k-1}\frac{2n+2-k}{(2n+2-k)^2+k^2}-\sum_{n=0}^{k-1}\frac{1}{(2n+2-k)^2+k^2} \\ \end{align} And now, since the original series was always real, we can extract the real part and leave the result unchanged! Therefore $$\boxed{\color{red}{S_1}=-\frac{1}{2k^2}-\frac{\pi}{2k}\frac{\sinh(k\pi)}{(-1)^k-\cosh(k\pi)}-\sum_{n=0}^{k-1}\frac{1}{(2n+2-k)^2+k^2}}$$ Additionally, since the result is real, then the imaginary part is $0$, and this leads to the first side-result: $$\boxed{\sum_{n=0}^{k-1}\frac{2n+2-k}{(2n+2-k)^2+k^2}=\frac{1}{2k}}$$ Now that we are done with $S_1$, let's attack $S_2$.
PART $2$: $S_2$
Remember that $$\color{blue}{S_2}=\frac{i}{2(k+1)}(\psi(z-\bar{z}k)-\psi(\bar{z}-zk))$$ As before, we will start by using the fact that $\bar{z}=1-z$: \begin{align} \psi(z-\bar{z}k) & =\psi(z-k+zk) \\ & \stackrel{\text{(4)}}{=}\psi(-z+k-zk)+\frac{1}{k-z-zk}+\pi\cot(\pi(k-z-zk)) \\ & \stackrel{\text{(3)}}{=}\psi(-z-zk)+\sum_{n=0}^{k-1}\frac{1}{n-z-zk}+\frac{1}{k-z-zk}+\pi\cot(\pi(k-z-zk)) \\ \end{align} while for the second digamma: \begin{align} \psi(\bar{z}-kz) & =\psi(1-z-zk) \\ & \stackrel{\text{(1)}}{=}\psi(-z-zk)-\frac{1}{z+zk} \\ \end{align} and therefore $$\psi(z-\bar{z}k)-\psi(\bar{z}-zk)=\frac{1}{k-z-zk}+\frac{1}{z+zk}+\sum_{n=0}^{k-1}\frac{1}{n-z-zk}+\pi\cot(\pi(k-z-zk))$$ Again, plug $z$ in and simplify: $$\frac{1}{k-z-zk}=\frac{2}{k-ik-(1+i)}=\frac{k-1}{k^2+1}+i\frac{k+1}{k^2+1}$$ $$\frac{1}{z+zk}=\frac{1}{1+k}-i\frac{1}{1+k}$$ $$\frac{1}{n-z-zk}=-\frac{k-2n+1}{n^2+(n-1)^2-2kn+k^2+2k}+i\frac{k+1}{n^2+(n-1)^2-2kn+k^2+2k} $$ $$\cot(\pi(k-z-zk))=-\frac{\sin(k\pi)}{\cos(k\pi)-\cosh((k+1)\pi)}+i\frac{\sinh((k+1)\pi)}{\cos(k\pi)-\cosh((k+1)\pi)}$$ and now, putting these values back in $S_2$, we end up with \begin{align} \color{blue}{S_2} & =\frac{i}{2(k+1)}(\psi(z-\bar{z}k)-\psi(\bar{z}-zk)) \\ & =\frac{i}{2}\frac{k-1}{(k+1)(k^2+1)}-\frac{1}{2(k^2+1)}+\frac{1+i}{2(k+1)^2}-\frac{i}{2(k+1)}\sum_{n=0}^{k-1}\frac{k-2n+1}{(k-n+1)^2+n^2}-\frac12\sum_{n=0}^{k-1}\frac{1}{(k-n+1)^2+n^2}-\frac{\pi}{2(k+1)}\frac{\sinh((k+1)\pi)}{(-1)^k-\cosh((k+1)\pi)} \end{align} but again, $S_2$ is real! So extracting the real part leads us to the final value for $S_2$: $$\boxed{\color{blue}{S_2}=\frac{1}{2(k+1)^2}-\frac{1}{2(k^2+1)}-\frac12\sum_{n=0}^{k-1}\frac{1}{(k-n+1)^2+n^2}-\frac{\pi}{2(k+1)}\frac{\sinh((k+1)\pi)}{(-1)^k-\cosh((k+1)\pi)}}$$ And, for the second side-result, just take the imaginary part and set it equal to $0$: $$\boxed{\sum_{n=0}^{k-1}\frac{k-2n+1}{(k-n+1)^2+n^2}=\frac{1}{k+1}+\frac{k-1}{k^2+1}}$$
Conclusion:
Finally, remember that we were after $\color{red}{S_1}+\color{blue}{S_2}$, and notice that \begin{align} \frac{\sinh(x)}{(-1)^k-\cosh(x)} & =-\coth\left(\frac{x}{2}\right) \hspace{1cm}\text{k even} \\ & =-\tanh\left(\frac{x}{2}\right) \hspace{1cm}\text{k odd} \\ \end{align} which shows why in the conjectured form there are only $\tanh$ and $\coth$. Furthermore, if I haven't made any typo, we now have an expression for the ratio $\frac{a_k}{b_k}$: \begin{align} \color{green}{\frac{a_k}{b_k}} & =\frac{1}{2(k+1)^2}-\frac{1}{2(k^2+1)}-\frac{1}{2k^2}-\frac12\sum_{n=0}^{k-1}\frac{1}{(k-n+1)^2+n^2}-\sum_{n=0}^{k-1}\frac{1}{(2n+2-k)^2+k^2} \\ \end{align} and I have checked that this agrees with the first few cases. This ugly representation also explains why the two sequences $a_k$, $b_k$ were so hard to guess. It remains to check if this is consistent with @Ydd's result for $\frac{a_k}{b_k}$ as $k\to\infty$, but honestly I can't see how to do it now.
Out of all of this, what I find most intriguing are the two closed forms for the sums obtained setting $\Im=0$.