$\color{brown}{\textbf{Sum representation.}}$
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.
$\color{brown}{\textbf{Results.}}$
- Closed form of the given sum has not obtained.
- Obtained alternative representation $(1).$
- Obtained calculation formulas $(2),(3),(6),(8)$ via fast convergent sums.
- In accordance with $(2)$ with 23 terms in the sums, the given sum equals to
$$S\approx \color{green}{\mathbf{0.92049\,17248\,05952\,11240\,27801\,93415\,98345\,40,}}$$
where all digits are correct.
- Increasing of the quantity of the first splitting terms in $(1)$ significantly accelerates convergence of the obtained series.
Best Answer
Note that $$ \begin{aligned} \int_{0}^{1} {\frac{x^{2n}}{1+x} \,\mathrm{d}x} & = \sum_{k=0}^{\infty} {(-1)^{k} \int_{0}^{1} {x^{2n+k} \,\mathrm{d}x}} = \sum_{k=0}^{\infty} {\frac{(-1)^{k}}{2n+k+1}} = \sum_{k=2n+1}^{\infty} {\frac{(-1)^{k+1}}{k}}\\ & = \sum_{k=1}^{\infty} {\frac{(-1)^{k+1}}{k}} - \sum_{k=1}^{2n} {\frac{(-1)^{k+1}}{k}} = \ln2 - H_{2n} + H_{n} \end{aligned} $$ so $$ \psi(n)-\psi(2n)=H_{n-1}-H_{2n-1}=H_n-H_{2n}-\frac1{2n}=\int_{0}^{1} {\frac{x^{2n}}{1+x} \,\mathrm{d}x} - \ln2 - \frac1{2n} $$ hence $$ \begin{aligned} &\, \sum_{n=1}^{\infty} \frac{(-1)^n}{n}(\psi(n)-\psi(2n)) \\ =&\, \int_{0}^{1}\left(\frac{1}{1+x}\sum_{n=1}^{\infty} {\frac{(-1)^nx^{2n}}{n}}\right) \mathrm{d}x-\ln2\sum_{n=1}^{\infty} {\frac{(-1)^n}{n}}-\frac1{2}\sum_{n=1}^{\infty} {\frac{(-1)^n}{n^2}}\\ =& -\int_{0}^{1} \frac{\ln(1+x^2)}{1+x}\,\mathrm{d}x + \ln^22 + \frac{\pi^2}{24} = \frac{\pi^2}{16} + \frac{\ln^22}{4} \end{aligned} $$ where $$ \int_{0}^{1} \frac{\ln(1+x^2)}{1+x}\,\mathrm{d}x = -\frac{\pi^2}{48} + \frac{3\ln^22}{4} $$ Supplement: to evaluate the last integral by elementary way, one is to denote $$ F(a)=\int_{0}^{1} \frac{\ln(1+a^2x^2)}{1+x}\,\mathrm{d}x $$ and $$ F'(a) = \int_{0}^{1} \frac{2ax^2}{(1+x)(1+a^2x^2)} \,\mathrm{d}x = -\frac{2\arctan a}{1+a^2} + \frac{2a\ln2}{1+a^2} + \frac{\ln(1+a^2)}{a(1+a^2)} $$ then antiderivative with $F(0)=0$ $$ \begin{aligned} F(a) & = \int_{0}^{a} \left(-\frac{2\arctan u}{1+u^2} + \frac{2u\ln2}{1+u^2} + \frac{\ln(1+u^2)}{u(1+u^2)}\right) \mathrm{d}u \\ & = -(\arctan a)^2 + \ln2 \ln(1+a^2) + \int_{0}^{a} \frac{\ln(1+u^2)}{u} \,\mathrm{d}u - \frac1{4}\ln^2(1+a^2) \end{aligned} $$ hence $$ \begin{aligned} \int_{0}^{1} \frac{\ln(1+x^2)}{1+x}\,\mathrm{d}x = F(1) & = -\frac{\pi^2}{16} + \frac{3\ln^22}{4} + \int_{0}^{1} \frac{\ln(1+u^2)}{u} \,\mathrm{d}u \\ & = -\frac{\pi^2}{16} + \frac{3\ln^22}{4} + \frac1{2} \int_{0}^{1} \frac{\ln(1+u)}{u} \,\mathrm{d}u \\ & = -\frac{\pi^2}{48} + \frac{3\ln^22}{4} \end{aligned} $$ Supplement 2: From this post, recalling that $$ \arctan x \ln (1+x^2) = 2\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} x^{2n+1}} $$ whose derivative gives you $$ \sum_{n=0}^{\infty} {(-1)^{n}H_{2n}x^{2n}} = -\frac{x\arctan x}{1+x^2} - \frac1{2}\frac{\ln(1+x^2)}{1+x^2} $$ which can help you finish $\omega_2$ from your approach. Yet lots of the terms will be cancelled by subtraction, there is no need to solve it term by term. Moreover, if we denote $$ H_{2n} = \sum_{k=0}^{n-1} \frac1{2n+1} + \frac{H_n}{2} $$ $$ \frac1{2}\sum_{n=0}^{\infty} {(-1)^{n}H_{n}x^{2n}} = -\frac1{2}\frac{\ln(1+x^2)}{1+x^2} $$ you can clearly find $$ \sum_{n=0}^{\infty} {(-1)^{n}\sum_{k=0}^{n-1} \frac{x^{2n}}{2n+1}} = -\frac{x\arctan x}{1+x^2} $$