$$
\begin{align}
&\sum_{n=2}^\infty(-1)^n\left(\log(2)-\frac1{n+1}-\frac1{n+2}-\cdots-\frac1{2n}\right)\\
&=\sum_{n=2}^\infty(-1)^n\left(\log(2)-1+\frac12-\frac13+\dots+\frac1{2n}\right)\\
&=\sum_{n=2}^\infty(-1)^n\left(\frac1{2n+1}-\frac1{2n+2}+\frac1{2n+3}-\dots\right)\\
&=\sum_{n=2}^\infty(-1)^n\left(\frac1{(2n+1)(2n+2)}+\frac1{(2n+3)(2n+4)}+\dots\right)\\
&=\sum_{n=1}^\infty\frac1{(4n+1)(4n+2)}\\
&=\frac14\sum_{n=1}^\infty\left(\frac1n-\frac1{n+\frac12}\right)
-\frac14\sum_{n=1}^\infty\left(\frac1n-\frac1{n+\frac14}\right)\\