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.
$$\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2-4} = \frac{3 \ln (2)}{16} – \frac{\pi^2+1}{8} – \frac{\pi}{2} \int_{0}^{\infty} \sin(4\pi x) (\psi (x) – \ln (x)) \, dx$$
Best Answer
$\color{brown}{\textbf{Sum representation.}}$
The given sum allows the transformations of $$S=\sum_{k=3}^\infty \dfrac{\ln k}{k^2-4} =\dfrac14\,\sum_{k=3}^\infty \ln k\left(\dfrac1{k-2} - \dfrac1{k+2}\right)$$ $$=\dfrac14\,\sum_{k=3}^\infty \ln k\left(\dfrac1{k-2} - \dfrac1{k-1}+\dfrac1{k-1} - \dfrac1{k} + \dfrac1{k} - \dfrac1{k+1} + \dfrac1{k+1} - \dfrac1{k+2}\right)$$ $$=\dfrac14\,\sum_{k=3}^\infty \ln k\left(\dfrac1{(k-2)(k-1)}+\dfrac1{(k-1)k} + \dfrac1{k(k+1)} + \dfrac1{(k+1)(k+2)}\right)$$ $$=\dfrac14\sum_{k=2}^\infty\left(\dfrac{\ln(k+1)}{k(k-1)} +\dfrac{\ln(k+1)}{k(k+1)}+\dfrac{\ln(k-1)}{k(k-1)} + \dfrac{\ln(k-1)}{k(k+1)}\right) -\dfrac1{24}\ln2 -\dfrac1{48}\ln2$$ $$=-\dfrac1{16}\ln2+\dfrac14\sum_{k=2}^\infty\,\ln(k^2-1)\left(\dfrac1{k(k-1)}+\dfrac1{k(k+1)}\right),$$ $$S=-\dfrac1{16}\ln2+\dfrac12\sum_{k=2}^\infty\,\dfrac{\ln(k^2-1)}{k^2-1}.\tag1$$ Representation $(1)$ looks more hard, than the given one. However, it can be splitted to the simple terms and convenient sums.
$\color{brown}{\textbf{Splitting.}}$
Some first terms of the obtained sum can be accounted apartly. Besides, the sum can be splitted to the pair of the convenient parts. So on, $$S=S_0 +S_1+S_2,\tag2$$ where $$S_0 = -\dfrac1{16}\ln2 + \dfrac16\ln3 +\dfrac1{16}\ln8+\dfrac1{30}\ln15+\dfrac1{48}\ln24,$$ $$S_0 =\dfrac1{240}(45\ln2+53\ln3+8\ln5)\approx0.42622\,32405\,17000\,64818\,55396\,57034\,7,\tag3$$ $$S_1=\dfrac12\sum_{k=6}^\infty\,\dfrac{\ln(k^2-1)-\ln k^2}{k^2-1},\quad S_2=\sum_{k=6}^\infty\,\dfrac{\ln k}{k^2-1}.\tag4$$
$\color{brown}{\textbf{The first sum.}}$
Is known the series representation (when $\;|x|<1\,$) in the form of $$\dfrac{x\ln(1-x)}{1-x} = -x^2-\dfrac32x^3-\dfrac{11}6x^4-\dots-\operatorname H_{j-1} x^j-\dots = -\sum\limits_{j=2}^\infty \operatorname H_{j-1}\,x^j,\tag5$$ where $\;\operatorname H_j=1+\frac12+\dots+\frac1j\;$ is a harmonic number.
Since $\;k\ge 6,\;$ then $$2S_1=\sum\limits_{k=6}^\infty \dfrac{\ln\left(1-\dfrac1{k^2}\right)}{k^2-1} =\sum\limits_{k=6}^\infty \dfrac{\dfrac1{k^2}\ln\left(1-\dfrac1{k^2}\right)}{1-\dfrac1{k^2}} =\sum\limits_{k=6}^\infty \left(-\sum\limits_{j=2}^\infty\dfrac{\operatorname H_{j-1}}{k^{2j}}\right),$$ $$S_1 = \dfrac12\sum\limits_{j=2}^\infty\,\operatorname H_{j-1} \left(1+\dfrac1{4^j}+\dfrac1{9^j}+\dfrac1{16^j}+\dfrac1{25^j}-\zeta(2j)\right),\tag6$$ where $\;\zeta(m) = 1+2^{-m}+3^{-m}+4^{-m}+\dots\;$ is the Riemann zeta-function.
Easily to see that the additional terms in the braces significanly accelerate the series convergence. In the choosen case, the first twenty terms of the sum in $(6)$ give $$S_1\approx -0.00101\,51087\,76078\,79322\,37019\,23747\,0,\tag7$$ where all digits are correct.
$\color{brown}{\textbf{The second sum.}}$
Taking in account OP summation formulas, similarly to previous one can get $$S_2=\sum_{k=6}^\infty\,\dfrac{\ln k}{k^2}\dfrac1{1-\dfrac1{k^2}} =\sum_{k=6}^\infty\,\ln k\sum\limits_{j=1}^\infty\dfrac1{k^{2j}},$$ $$S_2=\sum_{j=1}^\infty\,\left(-\zeta'(2j)-\dfrac{\ln2}{4^j} -\dfrac{\ln3}{9^j}-\dfrac{\ln4}{16^j}-\dfrac{\ln5}{25^j}\right).\tag8$$ The first twenty terms in $(8)$ give $$S_2\approx 0.49528\,35930\,65030\,25744\,09424\,60128\,3,\tag9$$ where all digits are correct.
Note that additional terms in the braces of $(8)$ provide fast convergence.
$\color{brown}{\textbf{Results.}}$