Playing around with the harmonic number $H(n) = 1+ \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$ and its asymptotic approximation
$$H(n\to\infty) \simeq \gamma + \log(n) +\frac{1}{2n} -\frac{1}{12 n^2} + ...$$
I started with first order comparisions studying these sums
$$s_H = \sum _{k=1}^{\infty } \frac{(-1)^k H_{k-1}}{k}$$ $$s_L= \sum _{k=1}^{\infty } \frac{(-1)^k (\log (k)+\gamma )}{k}$$
Notice the shift in the argument of $H(k)$.
Mathematica tells me that
$$s_H = \frac{1}{2}\log ^2(2)$$ $$s_L = - \frac{1}{2}\log ^2(2)$$
Normally we would expect that the sum using $H$ should be comparable to that using an approximation of $H$. But this expectation is not justified at all here.
Hence we are "naturally" led to ask for this sum
$$s_{HL} = \sum _{k=1}^{\infty } \frac{(-1)^k \left(H_{k-1}+\left(\log (k)+\gamma \right)\right)}{k}$$
Here Mathematica finds that
$$s_{HL} = 0\tag{*}$$
I wouldn't have guessed that from looking at the summands.
Hence an independent proof of $(*)$ would be desirable.
Remark: this problem is related to A closed form of the family of series $\sum _{k=1}^{\infty } \frac{\left(H_k\right){}^m-(\log (k)+\gamma )^m}{k}$ for $m\ge 1$
\begin{align} s_H&=\sum_{k=1}^\infty\frac{(-1)^kH_{k-1}}{k}\\ &=\sum_{k=1}^\infty\frac{(-1)^kH_{k}}{k}-\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\\ &=\frac12\ln^2(2)+\operatorname{Li}_2(-1)-\operatorname{Li}_2(-1)\\ &=\frac12\ln^22 \end{align} where we used the generating function $\ \sum_{k=1}^\infty\frac{x^k H_k}{k}=\frac12\ln^2(1-x)+\operatorname{Li}_2(x)$
.
From this solution , we proved $\ \int_0^\infty \ln x\ e^{-kx}\ dx=-\frac{\ln k+\gamma}{k}$ , so we can write
\begin{align} s_L&=\sum_{k=1}^\infty\frac{(-1)^k(\ln k+\gamma)}{k}\\ &=-\int_0^\infty\ln x\sum_{k=1}^\infty(-e^{-x})^k\ dx\\ &=\int_0^\infty\frac{e^{-x}\ln x}{1+e^{-x}}\ dx \end{align}
which I think manageable to prove it $-\frac12\ln^22$