How to evaluate $\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$? Is it possible to evaluate the sum:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$$
I expect it may be related to $\zeta^{\prime} (2)$:
$$\zeta^{\prime} (2) = - \sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$$
Is there an identity that works for my series, involving the natural logarithm, that is similar to the identity that:
$$\sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)} = \frac{\psi(a) - \psi(b)}{a-b}$$
Also potentially related, the Lüroth analogue of Khintchine’s constant can be defined as the following:
$$\sum_{n=1}^{\infty} \frac{\ln (n)}{n(n+1)}$$
as mentioned here.
After some work, the following can be shown:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{5\ln(2) + 4\ln(3)}{16} + \frac{1}{2} \sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right)$$
and furthermore:
$$\sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right) = \int_{0}^{2} \left( \frac{\left(1-\pi x \cot(\pi x) \right)}{2x^2} + \frac{1}{x^2 - 1} + \frac{1}{x^2 -4} \right) \, dx$$
EDIT
I have derived yet another form for my sum of interest, however, I found this one interesting as it seems like it could potentially be solvable?
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \int_{0}^{\infty} \left( \frac{\psi^{(0)} (s+3) + \gamma}{(s+2)(s-2)} - \frac{25}{16 (s-2)(s+1)} \right) \, ds$$
From this, it is possible to obtain the following:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4} i + \frac{25}{48} (\ln (2) - i \pi) - \frac{1}{8} + \frac{1}{16} i \pi + \frac{1}{4} \int_{0}^{i \pi} \psi^{(0)} \left( \frac{4}{1+ e^{u}} \right)  \, du$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4}i+\frac{25}{48} (\ln (2)-i \pi )+\frac{7 i \pi }{48}-\frac{1}{8}-\frac{\ln (2)}{3} -2 \int_0^{\infty } \frac{t \ln (\Gamma (1-i t))}{\left(t^2+4\right)^2} \, dt$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = -\frac{1}{8}-\frac{i \pi }{4}+\frac{i \gamma  \pi }{4}-\frac{\ln (2)}{16} - 2 \int_{0}^{\infty} \frac{t \ln (\Gamma (-i t)) }{(4+t^2)^2} \, dt$$
$$\implies \sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} =\frac{25}{48} \ln (2) -\frac{1}{8} + \int_{1}^{\infty} \frac{\ln (v-1) \text{li} (v^2)}{v^5} \, dv$$
Where $\text{li}$ is the logarithmic integral function.
$$\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2-4} = \frac{3 \ln (2)}{16} - \frac{\pi^2+1}{8} - \frac{\pi}{2} \int_{0}^{\infty} \sin(4\pi x) (\psi (x) - \ln (x)) \, dx$$

Considering the more general case $$S_a=\sum_{k=a+1}^{\infty} \frac{\log(k)}{k^2 - a^2}$$
$$\frac{\log (k)}{k^2 - a^2}=\sum_{n=0}^\infty \frac{\log (k)}{k^{2n+2}} a^{2n}$$
$$\sum_{k=a+1}^{\infty} \frac{\log(k)}{k^{2n+2}}=-\text{HurwitzZeta}^{(1,0)}(2 n+2,a+1)$$
$$S_a=-\sum_{n=0}^\infty  a^{2n}\,\text{HurwitzZeta}^{(1,0)}(2 n+2,a+1)$$ converges very fast.
If the summation had started at $k=1$ instead of $k=a+1$, we would have obtained
$$-\sum_{n=0}^\infty  a^{2n}\,\zeta '(2 (n+1))$$
Considering the case of $a=2$ and the partial sums up to $p$, the value $0.920492$ is obtained for $p=17$.
$$\left(
\begin{array}{cc}
 p & -\sum_{n=0}^p \\
 1 & 0.866620 \\
 2 & 0.898968 \\
 3 & 0.911406 \\
 4 & 0.916559 \\
 5 & 0.918769 \\
 6 & 0.919732 \\
 7 & 0.920155 \\
 8 & 0.920343 \\
 9 & 0.920426 \\
 10 & 0.920462 \\
 11 & 0.920479 \\
 12 & 0.920486 \\
 13 & 0.920489 \\
 14 & 0.920491 \\
 15 & 0.920491 \\
 16 & 0.920491 \\
 17 & 0.920492
\end{array}
\right)$$ while the orginal summation requires tens of thousands.
This is explained by the fact that
$$\zeta '(2 (n+1)) \sim -\exp\Big[-\frac{1106 }{797}n-\frac{713}{430} \Big]$$
