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.

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.
 A: $A(1,1)$:
$$
\begin{align}
\sum_{n=1}^N\frac{(-1)^{n-1}}{n}H_n
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\sum_{n=2}^N\frac{(-1)^{n-1}}{n}H_{n-1}\\
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{n=2}^N\sum_{k=1}^{n-1}\frac{(-1)^{n-1}}{n}\left(\frac1k+\frac1{n-k}\right)\\
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{n=2}^N\sum_{k=1}^{n-1}\frac{(-1)^{n-1}}{k(n-k)}\\
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{k=1}^{N-1}\sum_{n=k+1}^N\frac{(-1)^{n-1}}{k(n-k)}\\
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{k=1}^{N-1}\sum_{n=1}^{N-k}\frac{(-1)^{n+k-1}}{kn}\\
&=\color{#00A000}{\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}}
-\color{#0000FF}{\frac12\sum_{k=1}^{N-1}\frac{(-1)^{k-1}}{k}\sum_{n=1}^{N-1}\frac{(-1)^{n-1}}{n}}\\
&+\color{#C00000}{\frac12\sum_{k=1}^{N-1}\frac{(-1)^{k-1}}{k}\sum_{n=N-k+1}^{N-1}\frac{(-1)^{n-1}}{n}}\tag{1}
\end{align}
$$
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: $$\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}H_k=\sum_{k=1}^\infty (-1)^{k+1}H_k\int_0^1 x^{k-1}dx\\=\int_0^1\frac1x\sum_{k=1}^\infty{-H_k (-x)^{k}}dx=\int_0^1\frac{\ln(1+x)}{x(1+x)}dx\\=\int_0^1\frac{\ln(1+x)}{x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx\\=-\operatorname{Li}_2(-1)-\frac12\ln^22\\=\frac12\zeta(2)-\frac12\ln^22$$
where we used the identity $\sum_{n=1}^\infty H_nx^n=-\frac{\ln(1-x)}{1-x}$ and the value $\operatorname{Li}_2(-1)=-\frac12\zeta(2)$
A: A full derivation of $A(m,1), \ m\ge2$, is found in this answer,
\begin{equation*}
\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(m)}}{n}=\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x}{2}\right)}{1-x}\textrm{d}x
\end{equation*}
\begin{equation*}
=\frac{1}{2}\biggr(m\zeta (m+1)-2\log (2) \left(1-2^{1-m}\right) \zeta (m)-\sum_{k=1}^{m-2} \left(1-2^{-k}\right)\left(1-2^{1+k-m}\right)\zeta (k+1)\zeta (m-k)\biggr),
\end{equation*}
where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}$ represents the $n$th generalized harmonic number of order $m$ and $\zeta$ denotes the Riemann zeta function.
Also, a full solution to the case
\begin{equation*}
\sum_{k=1}^{\infty} (-1)^{k-1} \frac{H_k}{k^{2n}}=\left(n+\frac{1}{2}\right)\eta(2n+1)-\frac{1}{2}\zeta(2n+1)-\sum_{k=1}^{n-1}\eta(2k)\zeta(2n-2k+1), \ n\ge1.
\end{equation*}
may be found in Cornel's new article here.
A: For $A(1,2)$:
Using $$\sum_{n=1}^\infty H_nx^n=-\frac{\ln(1-x)}{1-x}$$
replace $x$ with $-x$ then multiply both sides by $-\frac{\ln x}{x}$ and integrate between $0$ and $1$ and use the fact that $\int_0^1 -x^{n-1}\ln xdx=\frac{1}{n^2}$ we get
$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^2}=\int_0^1\frac{\ln x\ln(1+x)}{x(1+x)}dx$$
$$=\underbrace{\int_0^1\frac{\ln x\ln(1+x)}{x}dx}_{IBP}-\underbrace{\int_0^1\frac{\ln x\ln(1+x)}{1+x}dx}_{IBP}$$
$$=\int_0^1\frac{\operatorname{Li}_2(-x)}{x}dx+\frac12\int_0^1\frac{\ln^2(1+x)}{x}dx$$
$$=-\frac34\zeta(3)+\frac12\left(\frac14\zeta(3)\right)=\boxed{-\frac58\zeta(3)}$$

Proof of $\int_0^1\frac{\ln^2(1+x)}{x}dx$:
Proof 1:
Using the algebraic identity 
$$b^2=\frac12(a-b)^2+\frac12(a+b)^2-a^2$$
let $a=\ln(1-x)$ and $b=\ln(1+x)$ we have 
$$\int_0^1\frac{\ln^2(1+x)}{x}\ dx=\frac12\underbrace{\int_0^1\frac{\ln^2\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}=y}+\frac12\underbrace{\int_0^1\frac{\ln^2(1-x^2)}{x}\ dx}_{1-x^2=y}-\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ dx}_{1-x=y}\\=\int_0^1\frac{\ln^2y}{1-y^2}\ dy+\frac14\int_0^1\frac{\ln^2y}{1-y}\ dy-\int_0^1\frac{\ln^2y}{1-y}\ dy\\=\frac12\int_0^1\frac{\ln^2y}{1+y}\ dy-\frac14\int_0^1\frac{\ln^2y}{1-y}\ dy=\frac12\left(\frac32\zeta(3)\right)-\frac14(2\zeta(3))=\boxed{\frac14\zeta(3)}$$
Proof 2:
Using the generalization
$$\int_0^1\frac{\ln^n(1+x)}{x}dx=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)+\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$

For $A(2,1)$:
By Cauchy product we have 
$$-\ln(1-x)\operatorname{Li}_2(x)=\sum_{n=1}^\infty\left(\frac{2H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac3{n^3}\right)x^n$$
Set $x=-1$ and rearrange we get
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n}=3\operatorname{Li}_3(-1)-\ln(2)\operatorname{Li}_2(-1)-2\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^2}$$
$$=3\left(-\frac34\zeta(3)\right)-\ln(2)\left(-\frac12\zeta(2)\right)-2\left(-\frac58\zeta(3)\right)=\boxed{\frac12\ln(2)\zeta(2)-\zeta(3)}$$
A: Using integral representation:
$$
A(1,1)= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} H_n = -\int_0^1 \sum_{n=1}^\infty (-x)^n H_n \frac{\mathrm{d} x }{x}
$$
Now:
$$
  -\sum_{n=1}^\infty (-x)^n H_n = -\sum_{n=1}^\infty x^n \sum_{k=0}^{n-1} (-1)^k \frac{(-1)^{n-k}}{n-k} = -\sum_{n=0}^\infty (-x)^n \cdot \sum_{k=1}^\infty \frac{(-x)^k}{k}  = \frac{\log(1+x)}{1+x}
$$
Thus
$$
   A(1,1) = \int_0^1 \frac{\log(1+x)}{1+x} \frac{\mathrm{d}x}{x} = \left. \left(-\frac{1}{2} \log^2(1+x) - \operatorname{Li}_2(-x) \right)\right|_{x = 0}^{x=1} = -\frac{1}{2} \log^2(2) - \operatorname{Li}_2(-1)
$$
But $\operatorname{Li}_2(-1) = \sum_{k=1}^\infty \frac{(-1)^k}{k^2} = \left(2^{1-2}-1\right) \zeta(2) = -\frac{1}{2} \zeta(2)$. Thus$$
  A(1,1) = \frac{1}{2} \left( \zeta(2) - \log^2(2)\right)
$$
A: For convenience define, $$S(m,p)=\sum_{(a,b)\in \mathbb{N^2}}\frac{(-1)^{a+b}}{a^m(a+b)^p}$$
So that,
$$S(m,p)+A(m,p)=\eta(m+p)$$
Where $\eta$ is the dirichlet eta function
Now since, $$\sum_{j=1}^{k-1}\frac{1}{a^j(a+b)^{k-j}}=\frac{a}{ba^k}-\frac{a}{b(a+b)^k}-\frac{1}{(a+b)^k}$$
We get the reccurence relation,
$$\sum_{j=1}^{k-1}A(j,k-j)=k\eta(k)-\ln(2)\eta(k-1)-A(1,k-1)$$
From which we get the value of $A(1,1)$
A: Actually it suffices to know the generating function 
$$\sum_{k\geq 1}H^{(p)}_kx^k=\frac{\mathrm{Li}_p(x)}{1-x}$$
Upon integrating we obtain 
$$\sum_{k\geq 1}\frac{H^{(p)}_k}{k}x^k=\mathrm{Li}_{p+1}(x)+\int^x_0 \frac{\mathrm{Li}_p(t)}{1-t}\,d t$$
$$\sum_{k\geq 1}\frac{H_k}{k}x^k=\mathrm{Li}_{2}(x)+\frac{1}{2}\log^2(1-x)$$
$$\sum_{k\geq 1}\frac{H_k}{k}(-1)^k=-\frac{\pi^2}{12}+\frac{1}{2}\log^2(2)$$
A: $A(2,1)$:
$$
\begin{align}
\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n^{(2)}}{n}
&=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^3}+\sum_{n=1}^\infty(-1)^{n-1}\frac{H_{n-1}^{(2)}}{n}\tag{1}\\
&=\frac34\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\sum_{k=1}^{n-1}\frac1{k^2}\tag{2}\\
&=\frac34\zeta(3)+\sum_{k=1}^\infty\sum_{n=k+1}^\infty\frac{(-1)^{n-1}}{nk^2}\tag{3}\\
&=\frac34\zeta(3)+\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{k+n-1}}{(k+n)k^2}\tag{4}\\
&=\frac34\zeta(3)+\sum_{k=1}^\infty\sum_{n=1}^\infty(-1)^{k+n-1}\left(\frac1{k^2n}-\frac1{kn(k+n)}\right)\tag{5}\\[6pt]
&=\frac34\zeta(3)-\frac12\zeta(2)\log(2)+\frac14\zeta(3)\tag{6}\\[9pt]
&=\zeta(3)-\frac12\zeta(2)\log(2)\tag{7}
\end{align}
$$
Justification:
$(1)$: $H_n^{(2)}=\frac1{n^3}+H_{n-1}^{(2)}$
$(2)$: expand $H_{n-1}^{(2)}$
$(3)$: change order of summation
$(4)$: reindex $n\mapsto k+n$
$(5)$: $\frac1{(k+n)k^2}=\frac1{k^2n}-\frac1{kn(k+n)}$
$(6)$: $\sum\limits_{k=1}^\infty\sum\limits_{n=1}^\infty\frac{(-1)^{k+n}}{kn(k+n)}=\frac14\zeta(3)$ from $(5)$ and $(7)$ of this answer
$(7)$: addition
Note that this answer was taken from this answer.  There, it is shown, using the Euler Series Transformation, that
$$
A(2,1)=\sum_{n=1}^\infty\frac{H_n}{2^nn^2}\tag{8}
$$
A: Interestingly, $$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}H_{n}^{-}}{n} = \frac{\zeta(2)}{2} {\color{red}{+}} \frac{\log^{2} (2)}{2}$$ where $H_{n}^{-}$ are the alternating harmonic numbers defined as $$H_{n}^{-} = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} .$$
One way to show this is to notice that $$ \begin{align} \log (2) - H_{n}^{-}  &= \sum_{k=n+1}^{\infty} \frac{(-1)^{k-1}}{k} \\ &= (-1)^{n}\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k+n} \\ &= (-1)^{n} \sum_{k=1}^{\infty} (-1)^{k-1} \int_{0}^{1} x^{k+n-1} \ dx \\ &= (-1)^{n} \int_{0}^{1} x^{n}\sum_{k=1}^{\infty}(-1)^{k-1} x^{k-1} \ dx \\ &= (-1)^{n} \int_{0}^{1} \frac{x^{n}}{1+x} \ dx . \end{align}$$
Thus an integral representation of the alternating harmonic numbers is $$ H_{n}^{-} = \log (2) + (-1)^{n-1} \int_{0}^{1} \frac{x^{n}}{1+x} \ dx .$$
The integral on the right can be evaluated in terms of the digamma function, and you'll get a closed-form expression for the alternating harmonic numbers.
But getting back to evaluating that sum, 
$$ \begin{align} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}H_{n}^{-}}{n} &= \log(2) \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} + \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1} \frac{x^{n}}{1+x} \ dx \\ &= \log^{2} (2) + \int_{0}^{1} \frac{1}{1+x} \sum_{n=1}^{\infty} \frac{x^{n}}{n} \ dx \\ &= \log^{2} (2) - \int_{0}^{1} \frac{\log (1-x)}{1+x} \ dx \\  &=\log^{2} 2 - \int_{1/2}^{1} \frac{\log \big(1-(2t-1) \big)}{2t} \ 2 \ dt \\ &= \log^{2}(2) - \int_{1/2}^{1} \frac{\log \big(2(1-t) \big)}{t} \ dt \\ &= \log^{2}(2) - \int_{1/2}^{1} \frac{\log 2}{t} \ dt - \int_{1/2}^{1} \frac{\log (1-t)}{t} \ dt \\ &= \log^{2}(2) - \log^{2}(2) + \text{Li}_{2}(1) - \text{Li}_{2} \left( \frac{1}{2}\right) \\ &=  \zeta(2) - \frac{\zeta(2)}{2} + \frac{\log^{2} (2)}{2} \\ &= \frac{\zeta (2)}{2} + \frac{\log^{2} (2)}{2} . \end{align}$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k} \over k}:\ {\large ?}}$

\begin{align}&\color{#c00000}{%
\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}}
=\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k}
\int_{0}^{1}{1 - t^{k} \over 1 - t}\,\dd t
\\[3mm]&=\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k}\int_{0}^{1}
\ln\pars{1 - t}\pars{-kt^{k - 1}}\,\dd t
=-\int_{0}^{1}\ln\pars{1 - t}\sum_{k = 1}^{\infty}\pars{-t}^{k - 1}\,\dd t
\\[3mm]&=-\int_{0}^{1}{\ln\pars{1 - t} \over 1 + t}\,\dd t
=-\,\int_{0}^{1}{\ln\pars{t} \over 2 - t}\,\dd t
=-\,\int_{0}^{1/2}{\ln\pars{2t} \over 1 - t}\,\dd t
=-\,\int_{0}^{1/2}{\ln\pars{1 - t} \over t}\,\dd t
\\[3mm]&=\int_{0}^{1/2}{{\rm Li}_{1}\pars{t} \over t}\,\dd t
\end{align}
  where $\ds{{\rm Li}_{s}\pars{z}}$ is a
  PolyLogarithm Function and we'll use well known properties of them as explained in the above mentioned link.

Then,
$$
\color{#c00000}{%
\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}}
=\int_{0}^{1/2}{\rm Li}_{2}'\pars{t}\,\dd t
={\rm Li}_{2}\pars{\half} - {\rm Li}_{2}\pars{0}
=\color{#c00000}{{\rm Li}_{2}\pars{\half}}
$$

$\ds{{\rm Li}_{2}\pars{\half}}$ is given in the above mentioned link:
  \begin{align}&\color{#66f}{\large%
\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}}
={\pi^{2} \over 12} - \half\,\ln^{2}\pars{2}
=\color{#66f}{\large\half\bracks{\zeta\pars{2} - \ln^{2}\pars{2}}}
\end{align}

A: This is a special case of Kouba's https://arxiv.org/abs/1010.1842 equation
$$\sum_{n\ge 1}(-)^{n-1}\frac{H_{kn}}{n}
=
\frac{(k^2+1)\pi^2}{24k}-\frac12 \sum_{j=0}^{k-1}\log^2\left(2\sin\frac{(2j+1)\pi}{2k}\right)
$$
