Logarithmic series as $\sum_{n=3}^{\infty}(-1)^n\ln\!\left(1+\frac1{2n}\right) \!\ln \!\left(\frac{2n^2+n-6}{2n^2+n-10}\!\right)$

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 $T$.

Answering Q1.

One has

$$ \sum_{n=3}^{\infty}\ln\!\Big(1+\frac1{2n}\Big) \!\ln \!\left(\!\frac{2n^2+n-6}{2n^2+n-10}\!\right)=\ln\!\Big(\frac32\Big) \!\ln \!\Big(\frac76\Big)+\ln\!\Big(\frac54\Big) \!\ln \!\Big(\frac98\Big). \tag1 $$

Proof. One may write $$ \begin{align} &\sum_{n=3}^{\infty}\ln\!\Big(1+\frac1{2n}\Big) \!\ln \!\left(\!\frac{2n^2+n-6}{2n^2+n-10}\!\right) \\=&\sum_{n=3}^{\infty}\ln\!\Big(1+\frac1{2n}\Big) \!\ln \left[\frac{(2n-3)(n+2)}{(2n+5)(n-2)}\right] \\=&\sum_{n=3}^{\infty}\ln\!\Big(1+\frac1{2n}\Big) \!\ln \!\left[\frac{(2n-3)(2n+4)}{(2n+5)(2n-4)}\right] \\=&\sum_{n=3}^{\infty}\ln\!\Big(1+\frac1{2n}\Big) \!\left[\ln \!\left(\frac{2n-3}{2n-4}\right)-\ln\left(\frac{2n+5}{2n+4}\right)\right] \\=&\sum_{n=3}^{\infty}\ln\!\Big(1+\frac1{2n}\Big) \!\left[\ln \!\left(1+\frac{1}{2(n-2)}\right)-\ln\left(1+\frac{1}{2(n+2)}\right)\right] \\=&\sum_{n=3}^{\infty}\ln\!\Big(1+\frac1{2n}\Big) \!\ln \!\left(1+\frac{1}{2(n-2)}\right)-\sum_{n=3}^{\infty}\ln\!\Big(1+\frac1{2n}\Big)\ln\left(1+\frac{1}{2(n+2)}\right) \\=&\sum_{\color{red}{n=1}}^{\infty}\ln\!\Big(1+\frac1{2(n+2)}\Big) \!\ln \!\left(1+\frac{1}{2n}\right)-\sum_{\color{blue}{n=3}}^{\infty}\ln\!\Big(1+\frac1{2n}\Big)\ln\left(1+\frac{1}{2(n+2)}\right) \\=&\ln\!\Big(\frac32\Big) \!\ln \!\Big(\frac76\Big)+\ln\!\Big(\frac54\Big) \!\ln \!\Big(\frac98\Big), \end{align} $$ where we have made a change of index.

Similarly, one has

$$ \sum_{n=3}^{\infty}(-1)^n\ln\!\Big(1+\frac1{2n}\Big) \!\ln \!\left(\!\frac{2n^2+n-6}{2n^2+n-10}\!\right)=\ln\!\Big(\frac54\Big) \!\ln \!\Big(\frac98\Big)-\ln\!\Big(\frac32\Big) \!\ln \!\Big(\frac76\Big). \tag2 $$

Proof. One may write $$ \begin{align} &\sum_{n=3}^{\infty}(-1)^n\ln\!\Big(1+\frac1{2n}\Big) \!\ln \!\left(\!\frac{2n^2+n-6}{2n^2+n-10}\!\right) \\=&\sum_{n=3}^{\infty}(-1)^n\ln\!\Big(1+\frac1{2n}\Big) \!\ln \left[\frac{(2n-3)(n+2)}{(2n+5)(n-2)}\right] \\=&\sum_{n=3}^{\infty}(-1)^n\ln\!\Big(1+\frac1{2n}\Big) \!\ln \!\left[\frac{(2n-3)(2n+4)}{(2n+5)(2n-4)}\right] \\=&\sum_{n=3}^{\infty}(-1)^n\ln\!\Big(1+\frac1{2n}\Big) \!\left[\ln \!\left(\frac{2n-3}{2n-4}\right)-\ln\left(\frac{2n+5}{2n+4}\right)\right] \\=&\sum_{n=3}^{\infty}(-1)^n\ln\!\Big(1+\frac1{2n}\Big) \!\left[\ln \!\left(1+\frac{1}{2(n-2)}\right)-\ln\left(1+\frac{1}{2(n+2)}\right)\right] \\=&\sum_{n=3}^{\infty}(-1)^n\ln\!\Big(1+\frac1{2n}\Big) \!\ln \!\left(1+\frac{1}{2(n-2)}\right)-\sum_{n=3}^{\infty}(-1)^n\ln\!\Big(1+\frac1{2n}\Big)\ln\left(1+\frac{1}{2(n+2)}\right) \\=&\sum_{\color{red}{n=1}}^{\infty}(-1)^{n+\color{red}{2}}\ln\!\left(1+\frac1{2(n+2)}\right) \!\ln \!\left(1+\frac{1}{2n}\right)-\sum_{\color{blue}{n=3}}^{\infty}(-1)^n\ln\!\left(1+\frac1{2n}\right)\ln\left(1+\frac{1}{2(n+2)}\right) \\=&-\ln\!\Big(\frac32\Big) \!\ln \!\Big(\frac76\Big)+\ln\!\Big(\frac54\Big) \!\ln \!\Big(\frac98\Big), \end{align} $$ where we have made a change of index.

Answering Q2.

The given series are just an instance ($k=1$) of the following family.

Proposition. Let $k=1,2,3,\cdots.$ Then $$ \begin{align} {\small \sum_{n=2k+1}^{\infty}}&{\small \ln\!\left(\!1+\frac1{2n}\!\right) \!\ln \!\left(\!\frac{2n^2+n-8k^2+2k}{2n^2+n-8k^2-2k}\!\right)=\sum_{n=1}^{2k}\ln\!\left(\!1+\frac1{2n}\!\right) \!\ln\!\left(\!1+\frac1{2n+4k}\!\right)}, \tag3 \\\\ {\small \sum_{n=2k+1}^{\infty}(-1)^n} &{\small \ln\!\left(\!1+\frac1{2n}\!\right) \!\ln \!\left(\!\frac{2n^2+n-8k^2+2k}{2n^2+n-8k^2-2k}\!\right)=\sum_{n=1}^{2k}(-1)^n\ln\!\left(\!1+\frac1{2n}\!\right) \!\ln\!\left(\!1+\frac1{2n+4k}\!\right)}. \tag4 \end{align} $$