real analysis - Proving an alternating Euler sum: $sum_{k=1}^{infty} frac{(-1)^{k+1} H_k}{k} = frac{1}{2} zeta(2) - frac{1}{2} log^2 2$

Let $$A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},$$
where $H^{(p)}_n = \sum_{i=1}^n i^{-p}$, the $n$th $p$-harmonic number. The $A(p,q)$'s are known as alternating Euler sums.

Can someone provide a nice proof that
$$A(1,1) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2?$$

I worked for a while on this today but was unsuccessful. Summation by parts, swapping the order of summation, and approximating $H_k$ by $\log k$ were my best ideas, but I could not get any of them to work. (Perhaps someone else can?) I would like a nice proof in order to complete my answer here.

Bonus points for proving $A(1,2) = \frac{5}{8} \zeta(3)$ and $A(2,1) = \zeta(3) - \frac{1}{2}\zeta(2) \log 2$, as those are the other two alternating Euler sums needed to complete my answer.

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Added: I'm going to change the accepted answer to robjohn's $A(1,1)$ calculation as a proxy for the three answers he gave here. Notwithstanding the other great answers (especially the currently most-upvoted one, the one I first accepted), robjohn's approach is the one I was originally trying. I am pleased to see that it can be used to do the $A(1,1)$, $A(1,2)$, and $A(2,1)$ derivations.

Answer

$A(1,1)$:
$$
\begin{align}

\sum_{n=1}^N\frac{(-1)^{n-1}}{n}H_n
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\sum_{n=2}^N\frac{(-1)^{n-1}}{n}H_{n-1}\\
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{n=2}^N\sum_{k=1}^{n-1}\frac{(-1)^{n-1}}{n}\left(\frac1k+\frac1{n-k}\right)\\
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{n=2}^N\sum_{k=1}^{n-1}\frac{(-1)^{n-1}}{k(n-k)}\\
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{k=1}^{N-1}\sum_{n=k+1}^N\frac{(-1)^{n-1}}{k(n-k)}\\
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{k=1}^{N-1}\sum_{n=1}^{N-k}\frac{(-1)^{n+k-1}}{kn}\\
&=\color{#00A000}{\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}}
-\color{#0000FF}{\frac12\sum_{k=1}^{N-1}\frac{(-1)^{k-1}}{k}\sum_{n=1}^{N-1}\frac{(-1)^{n-1}}{n}}\\
&+\color{#C00000}{\frac12\sum_{k=1}^{N-1}\frac{(-1)^{k-1}}{k}\sum_{n=N-k+1}^{N-1}\frac{(-1)^{n-1}}{n}}\tag{1}
\end{align}

$$
where, using the Alternating Series Test, we have
$$
\begin{align}
&\color{#C00000}{\frac12\left|\sum_{k=1}^{N-1}\frac{(-1)^{k-1}}{k}\sum_{n=N-k+1}^{N-1}\frac{(-1)^{n-1}}{n}\right|}\\
&\le\frac12\left|\sum_{k=1}^{N/2}\frac{(-1)^{k-1}}{k}\sum_{n=N-k+1}^{N-1}\frac{(-1)^{n-1}}{n}\right|
+\frac12\left|\sum_{k=N/2}^{N-1}\frac{(-1)^{k-1}}{k}\sum_{n=N-k+1}^{N-1}\frac{(-1)^{n-1}}{n}\right|\\
&\le\frac12\cdot1\cdot\frac2N+\frac12\cdot\frac2N\cdot1\\
&=\frac2N\tag{2}
\end{align}

$$
Applying $(2)$ to $(1)$ and letting $N\to\infty$, we get
$$
\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}H_n=\color{#00A000}{\frac12\zeta(2)}-\color{#0000FF}{\frac12\log(2)^2}\tag{3}
$$