show that $$\sum_{n=1}^{\infty}\dfrac{(-1)^n\ln{n}}{n^2}+\sum_{n=1}^{\infty}\dfrac{(-1)^n\ln^2{n}}{n}=\dfrac{\pi^2}{12}\ln{2}+\ln^2{2}\left[\gamma-\dfrac{1}{3}\ln{2}\right]?$$
where the $\gamma$ is Eluer constant.
I remember this sum is old: $$\sum_{n=1}^{\infty}\dfrac{(-1)^n\ln{n}}{n^2}=\dfrac{\pi^2}{12}\ln{2}?$$ and think the general case $$\sum_{n=1}^{\infty}\dfrac{(-1)^n\ln^p{n}}{n^q}=?$$
I have found this :link use idea: since $$f(x)=\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n^x}$$ then $$f'(x)=-\sum_{n=1}^{\infty}\dfrac{(-1)^n\ln{n}}{n^x}$$ $$f''(x)=\sum_{n=1}^{\infty}\dfrac{(-1)^n\ln^2{n}}{n^x}$$ $$f'''(x)=\sum_{n=1}^{\infty}\dfrac{(-1)^n\ln^3{n}}{n^x}$$ $$\cdots$$ But Now How find this value $f'(1),f'(2),f''(1),f''(2),\cdots$?
maybe have some paper research it? Thank you