Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$
Note that $$\dfrac{(-1)^{k-1}}k = \int_0^1 (-x)^{k-1}dx$$ and $$\dfrac1n = \int_0^1 y^{n-1}dy$$
For the first one, \begin{align} \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}k \sum_{n=1}^k \dfrac1n & = \sum_{k=1}^{\infty} \sum_{n=1}^k \int_0^1 (-x)^{k-1}dx \int_0^1 y^{n-1} dy\\ & = \sum_{n=1}^{\infty} \sum_{k=n}^{\infty} \int_0^1 (-x)^{k-1}dx \int_0^1 y^{n-1} dy\\ & = \sum_{n=1}^{\infty} \int_0^1 \dfrac{(-x)^{n-1}}{1+x}dx \int_0^1 y^{n-1} dy\\ & = \int_0^1 \int_0^1\sum_{n=1}^{\infty} \dfrac{(-xy)^{n-1}}{1+x}dx dy\\ & = \int_0^1 \int_0^1\dfrac1{(1+x)(1+xy)}dx dy\\ & = \int_0^1 \int_0^1\dfrac1{(1+x)(1+xy)}dy dx\\ & = \int_0^1 \dfrac{\log(1+x)}{x(1+x)} dx\\ & = \int_0^1 \dfrac{\log(1+x)}{x} dx - \int_0^1 \dfrac{\log(1+x)}{(1+x)} dx\\ & = \dfrac{\zeta(2)}2 - \dfrac{\log^2 2}2 \end{align}
$$\int_0^1 \dfrac{\log(1+x)}{x} dx = \sum_{k=0}^{\infty} \int_0^1 \dfrac{(-1)^kx^k}{k+1} dx = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+1)^2} = \dfrac{\zeta(2)}2$$ $$\int_0^1 \dfrac{\log(1+x)}{(1+x)} dx = \left. \dfrac{\log^2(1+x)}2 \right \vert_{x=0}^{x=1} = \dfrac{\log^2 2}2$$
For the second one,
$$A(1,2) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \dfrac1n $$ $$\dfrac{(-1)^{k-1}}{k^2} = \int_0^1 (-x)^{k-1} dx \int_0^1 z^{k-1} dz = (-1)^{k-1} \int_0^1 \int_0^1 (xz)^{k-1} dx dz$$ \begin{align} \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \dfrac1n & = \sum_{k=1}^{\infty} \sum_{n=1}^k \int_0^1\int_0^1 (-1)^{k-1} (xz)^{k-1}dxdz \int_0^1 y^{n-1} dy\\ & = \int_0^1 \int_0^1 \int_0^1 \sum_{n=1}^{\infty} \dfrac{(-xyz)^{n-1}}{1+xz} dx dy dz\\ & = \int_0^1 \int_0^1 \int_0^1 \dfrac1{(1+xz)(1+xyz)} dx dy dz\\ & = \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz(1+xz)} dx dz\\ & = \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz} dx dz - \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{1+xz} dx dz\\ & = \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz} dx dz- \int_0^1 \dfrac{\log^2(1+z)}{2z} dz\\ & = \dfrac34 \zeta(3) - \dfrac{\zeta(3)}8\\ & = \dfrac58 \zeta(3) \end{align}
$$ \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz} dx dz = \sum_{k=0}^{\infty} \int_0^1 \int_0^1 \dfrac{(-1)^k (xz)^k}{k+1} dx dz = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+1)^3} = \dfrac34 \zeta(3)$$
For the third one, $$A(2,1) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k} \sum_{n=1}^k \dfrac1{n^2} $$ \begin{align} \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k} \sum_{n=1}^k \dfrac1{n^2} & = \int_0^1 \int_0^1 \int_0^1 \sum_{k=1}^{\infty} \sum_{n=1}^k (-1)^{k-1} x^{k-1} (yz)^{n-1} dx dy dz\\ & = \int_0^1 \int_0^1 \int_0^1 \sum_{n=1}^{\infty} \sum_{k=n}^{\infty} (-1)^{k-1} x^{k-1} (yz)^{n-1} dx dy dz\\ & = \int_0^1 \int_0^1 \int_0^1 \sum_{n=1}^{\infty} \dfrac{(-xyz)^{n-1}}{1+x} dx dy dz\\ & = \int_0^1 \int_0^1 \int_0^1 \dfrac1{(1+x)(1+xyz)} dx dy dz\\ & = \int_0^1 \int_0^1 \dfrac{\log(1+xy)}{(1+x)(xy)} dx dy\\ & = \zeta(3) - \dfrac{\zeta(2) \log 2}2 \end{align}
In general, if I have not made any mistake, this can be extended to $A(p,q)$. $$A(p,q) = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q}}{(1+x_1 x_2 \cdots x_q)(1+x_1 x_2 \cdots x_{p+q})}$$
Proceeding along similar lines, we also get that $$B(p,q) = \sum_{k=1}^{\infty} \dfrac{H_k^{(p)}}{k^q} = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q}}{(1-x_1 x_2 \cdots x_q)(1-x_1 x_2 \cdots x_{p+q})}$$
We also get that $$C(p,q) = \sum_{k=1}^{\infty} \dfrac1{k^q} \sum_{i=1}^k \dfrac{(-1)^{i-1}}{i^p} = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q}}{(1-x_1 x_2 \cdots x_q)(1+x_1 x_2 \cdots x_{p+q})}$$ $$D(p,q) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k^q} \sum_{i=1}^k \dfrac{(-1)^{i-1}}{i^p} = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q}}{(1+x_1 x_2 \cdots x_q)(1-x_1 x_2 \cdots x_{p+q})}$$
By the same argument as above, in general, nested sums like $$\sum_{k=1}^{\infty} \dfrac{(\pm 1)^{k-1}}{k^q} \sum_{n=1}^k \dfrac{(\pm 1)^{n-1}}{n^p} \sum_{m=1}^n \dfrac{(\pm 1)^{m-1}}{m^r} \cdots $$ equals $$\underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q+r+\cdots \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q+r+\cdots}}{(1\mp x_1 \cdots x_q)(1(\mp)(\pm)x_1 \cdots x_{p+q}) \cdots (1(\mp)(\pm)\cdots(\pm)x_1 \cdots x_{p+q+r+\cdots})}$$
For instance, $$\sum_{k=1}^{\infty} \dfrac{1}{k^q} \sum_{n=1}^k \dfrac{1}{n^p} \sum_{m=1}^n \dfrac{1}{m^r} = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q+r \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q+r}}{(1- x_1 \cdots x_q)(1-x_1 \cdots x_{p+q}) \cdots (1-x_1 \cdots x_{p+q+r})}$$ $$\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k^q} \sum_{n=1}^k \dfrac{1}{n^p} \sum_{m=1}^n \dfrac{1}{m^r} = \underbrace{\int_0^1 \cdots \int_0^1}_{p+q+r \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q+r}}{(1+ x_1 \cdots x_q)(1+x_1 \cdots x_{p+q}) \cdots (1+x_1 \cdots x_{p+q+r})}$$ $$\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k^q} \sum_{n=1}^k \dfrac{(-1)^{n-1}}{n^p} \sum_{m=1}^n \dfrac{1}{m^r} = \underbrace{\int_0^1 \cdots \int_0^1}_{p+q+r \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q+r}}{(1+ x_1 \cdots x_q)(1-x_1 \cdots x_{p+q}) \cdots (1-x_1 \cdots x_{p+q+r})}$$ $$\sum_{k=1}^{\infty} \dfrac{1}{k^q} \sum_{n=1}^k \dfrac{(-1)^{n-1}}{n^p} \sum_{m=1}^n \dfrac{1}{m^r} = \underbrace{\int_0^1 \cdots \int_0^1}_{p+q+r \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q+r}}{(1- x_1 \cdots x_q)(1+x_1 \cdots x_{p+q}) \cdots (1+x_1 \cdots x_{p+q+r})}$$
Similarly, for negative $p$,$q$ $r$ etc, we can replace the integrals $\int_0^1$ by the appropriate differentiation operator evaluated at $1$. I will post this in detail sometime over the weekend.
$A(1,1)$: $$ \begin{align} \sum_{n=1}^N\frac{(-1)^{n-1}}{n}H_n &=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\sum_{n=2}^N\frac{(-1)^{n-1}}{n}H_{n-1}\\ &=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{n=2}^N\sum_{k=1}^{n-1}\frac{(-1)^{n-1}}{n}\left(\frac1k+\frac1{n-k}\right)\\ &=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{n=2}^N\sum_{k=1}^{n-1}\frac{(-1)^{n-1}}{k(n-k)}\\ &=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{k=1}^{N-1}\sum_{n=k+1}^N\frac{(-1)^{n-1}}{k(n-k)}\\ &=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{k=1}^{N-1}\sum_{n=1}^{N-k}\frac{(-1)^{n+k-1}}{kn}\\ &=\color{#00A000}{\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}} -\color{#0000FF}{\frac12\sum_{k=1}^{N-1}\frac{(-1)^{k-1}}{k}\sum_{n=1}^{N-1}\frac{(-1)^{n-1}}{n}}\\ &+\color{#C00000}{\frac12\sum_{k=1}^{N-1}\frac{(-1)^{k-1}}{k}\sum_{n=N-k+1}^{N-1}\frac{(-1)^{n-1}}{n}}\tag{1} \end{align} $$ 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 $$ \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}H_n=\color{#00A000}{\frac12\zeta(2)}-\color{#0000FF}{\frac12\log(2)^2}\tag{3} $$
$A(1,2)$: $$ \begin{align} \sum_{n=1}^\infty\frac1{n^2}H_n &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{n^2}\left(\frac1k-\frac1{k+n}\right)\\ &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{nk(k+n)}\tag{1}\\ &=\sum_{k=1}^\infty\sum_{n=k+1}^\infty\frac1{nk(n-k)}\\ &=\sum_{n=2}^\infty\sum_{k=1}^{n-1}\frac1{nk(n-k)}\\ &=\sum_{n=2}^\infty\sum_{k=1}^{n-1}\frac1{n^2}\left(\frac1k+\frac1{n-k}\right)\\ &=2\sum_{n=1}^\infty\frac1{n^2}H_{n-1}\\ &=2\sum_{n=1}^\infty\frac1{n^2}H_n-2\zeta(3)\tag{2}\\ \sum_{n=1}^\infty\frac1{n^2}H_n &=2\zeta(3)\tag{3} \end{align} $$ $$ \begin{align} \sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_n &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^n}{n^2}\left(\frac1k-\frac1{k+n}\right)\\ &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^n}{nk(k+n)}\tag{4}\\ \sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_n &=-\frac34\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_{n-1}\\ &=-\frac34\zeta(3)+\frac12\sum_{n=1}^\infty\sum_{k=1}^{n-1}\frac{(-1)^n}{n^2}\left(\frac1k+\frac1{n-k}\right)\\ &=-\frac34\zeta(3)+\frac12\sum_{k=1}^\infty\sum_{n=k+1}^\infty\frac{(-1)^n}{nk(n-k)}\\ &=-\frac34\zeta(3)+\frac12\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{n+k}}{(n+k)kn}\tag{5} \end{align} $$ Using $\color{#C00000}{(1)}$, $\color{#C00000}{(3)}$, $\color{#00A000}{(4)}$, $\color{#0000FF}{(4)}$, and $\color{#C0A000}{(5)}$ along with the fact that $1+(-1)^k+(-1)^n+(-1)^{n+k}=4$ iff $k$ and $n$ are both even and $0$ otherwise: $$ \begin{align} \zeta(3) &=\frac12\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{nk(n+k)}\\ &=\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{\color{#C00000}{1}+\color{#00A000}{(-1)^k}+\color{#0000FF}{(-1)^n}+\color{#C0A000}{(-1)^{n+k}}}{nk(n+k)}\\ &=\color{#C00000}{2\zeta(3)}+\color{#00A000}{\sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_n}+\color{#0000FF}{\sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_n} +\color{#C0A000}{2\sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_n+\frac32\zeta(3)}\\ \hspace{-8mm}-\frac58\zeta(3) &=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_n\tag{6} \end{align} $$ That is, $$ \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}H_n=\frac58\zeta(3)\tag{7} $$