Logarithmic series as $\sum_{n=3}^{\infty}(-1)^n\ln\!\left(1+\frac1{2n}\right) \!\ln \!\left(\frac{2n^2+n-6}{2n^2+n-10}\!\right)$ Inspired by this question, I've designed the following series. 

$$
\begin{align}
S&=\sum_{n=3}^{\infty}\ln\!\left(1+\frac1{2n}\right) \!\ln \!\left(\frac{2n^2+n-6}{2n^2+n-10}\!\right) \tag1
\\\\
T&=\sum_{n=3}^{\infty}(-1)^n\ln\!\left(1+\frac1{2n}\right) \!\ln \!\left(\frac{2n^2+n-6}{2n^2+n-10}\!\right) \tag2
\end{align}
$$

Each one admits a nice closed form.
Q1. What are their closed forms?
Q2. To which family would you tell these series belong to ?
 A: Noting that
$$ 2n^2+n-6=(n+2)(2n-3),2n^2+n-10=(n-2)(2n+5)$$
one has
\begin{eqnarray}
S&=&\sum_{n=3}^{\infty}\ln\!\left(1+\frac1{2n}\right) \!\ln \!\left(\frac{2n^2+n-6}{2n^2+n-10}\!\right)\\
&=&\sum_{n=3}^{\infty}\ln\!\left(\frac{2n+1}{2n}\right) \!\ln \!\left(\frac{(n+2)(2n-3)}{(n-2)(2n+5)}\!\right)\\
&=&-\sum_{n=3}^{\infty}\ln\!\left(\frac{2n+1}{2n}\right) \!\ln \!\left(\frac{(2n+5)(2n-4)}{(2n-3)(2n+4)}\!\right)\\
&=&-\sum_{n=3}^{\infty}\ln\!\left(\frac{2n+1}{2n}\right) \!\left(\ln\frac{2n+5}{2n+4}-\ln\frac{2n-3}{2n-4}\right)\\
&=&-\sum_{n=3}^{\infty}\left[\ln\left(\frac{2n+1}{2n}\right) \!\left(\ln\left(\frac{2n+5}{2n+4}\right)-\ln\left(\frac{2n-3}{2n-4}\right)\right)\right]\\
&=&-\sum_{n=3}^{\infty}\left[\ln\left(\frac{2n+5}{2n+4}\right)\ln\left(\frac{2n+1}{2n}\right) -\ln\left(\frac{2n+1}{2n}\right)ln\left(\frac{2n-3}{2n-4}\right)\right]\\
&=&-\bigg[\ln\left(\frac{11}{10}\right)\ln\left(\frac{7}{6}\right)-\ln\left(\frac{7}{6}\right)\ln\left(\frac{3}{2}\right)+\ln\left(\frac{13}{12}\right)\ln\left(\frac{9}{8}\right)-\ln\left(\frac{9}{8}\right)\ln\left(\frac{5}{4}\right)\\
&&+\ln\left(\frac{15}{14}\right)\ln\left(\frac{11}{10}\right)-\ln\left(\frac{11}{10}\right)\ln\left(\frac{7}{6}\right)+\ln\left(\frac{19}{18}\right)\ln\left(\frac{15}{14}\right)-\ln\left(\frac{15}{14}\right)\ln\left(\frac{11}{10}\right)\\
&&+\ln\left(\frac{23}{22}\right)\ln\left(\frac{19}{18}\right)-\ln\left(\frac{19}{18}\right)\ln\left(\frac{15}{14}\right)+\cdots\bigg]\\
&=&\ln\left(\frac{7}{6}\right)\ln\left(\frac{3}{2}\right)+\ln\left(\frac{9}{8}\right)\ln\left(\frac{5}{4}\right).
\end{eqnarray}
One can use the same way to handle $T$.
