$$\sum_{k=1}^{\infty} (-1)^{k}\frac{(\ln{k})^{2}}{k} \space\space\space ?$$
From $\sum_{n\geq 1}\frac{(-1)^n \ln n}{n}$ it has been answered that
$$ \sum_{k=1}^{\infty} (-1)^{k} \frac{\ln{k}}{k} = \space \gamma \cdot \ln{2} \space - \space \frac{{(\ln{2})^{2}}}{2} \space \space \space , $$
where $\gamma$ is the Euler-Mascheroni constant.
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Hint. From the standard relation $$\sum_{n\geq 1}\frac{(-1)^n }{n^s}=\left(\frac{1}{2^{s-1}}-1\right)\zeta(s)$$ one may differentiate twice and set $s \to 1^+$ using the Laurent series of the Riemann zeta function.