How to prove this series: $\displaystyle \sum_{n=1}^{\infty }\frac{\left ( -1 \right )^{n}\ln n}{n}=\gamma \ln 2-\frac{1}{2}\ln^22$

Question

How to prove this series $$\sum_{n=1}^{\infty }\frac{\left ( -1 \right )^{n}\ln n}{n}=\gamma \ln 2-\frac{1}{2}\ln^22$$ and \begin{align*} \sum_{n=1}^{\infty }\frac{\left ( -1 \right )^{n}\ln \left ( n+1 \right )}{n+1}&=\frac{1}{2}\ln^22-\gamma \ln 2\\ \sum_{n=1}^{\infty }\frac{\left ( -1 \right )^{n}\ln \left ( n+2 \right )}{n+2}&=\gamma \ln 2-\frac{1}{2}\ln^22-\frac{1}{2}\ln 2 \end{align*} So I want to know, is there a closed form for

$$\sum_{n=1}^{\infty }\frac{\left ( -1 \right )^{n}\ln \left ( n+k \right )}{n+k}~~~\left ( k>0 \right )$$

Answer 1

We have, for any $\alpha>1$: $$ \sum_{n\geq 1}\frac{(-1)^n}{n^\alpha} = (2^{1-\alpha}-1)\cdot\zeta(\alpha) \tag{1}$$ hence the main result (or the secondary one up to a shift of the summation variable) follows from considering the opposite of the derivative with respect to $\alpha$ of both sides, followed by an evaluation of the limit for $\alpha\to 1^+$.