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

Question

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 1

Note that $$\dfrac{(-1)^{k-1}}k = \int_0^1 (-x)^{k-1}dx$$ and $$\dfrac1n = \int_0^1 y^{n-1}dy$$

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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}

$$\int_0^1 \dfrac{\log(1+x)}{x} dx = \sum_{k=0}^{\infty} \int_0^1 \dfrac{(-1)^kx^k}{k+1} dx = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+1)^2} = \dfrac{\zeta(2)}2$$ $$\int_0^1 \dfrac{\log(1+x)}{(1+x)} dx = \left. \dfrac{\log^2(1+x)}2 \right \vert_{x=0}^{x=1} = \dfrac{\log^2 2}2$$

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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}

$$ \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz} dx dz = \sum_{k=0}^{\infty} \int_0^1 \int_0^1 \dfrac{(-1)^k (xz)^k}{k+1} dx dz = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+1)^3} = \dfrac34 \zeta(3)$$

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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}

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In general, if I have not made any mistake, this can be extended to $A(p,q)$. $$A(p,q) = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q}}{(1+x_1 x_2 \cdots x_q)(1+x_1 x_2 \cdots x_{p+q})}$$

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Proceeding along similar lines, we also get that $$B(p,q) = \sum_{k=1}^{\infty} \dfrac{H_k^{(p)}}{k^q} = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q}}{(1-x_1 x_2 \cdots x_q)(1-x_1 x_2 \cdots x_{p+q})}$$

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We also get that $$C(p,q) = \sum_{k=1}^{\infty} \dfrac1{k^q} \sum_{i=1}^k \dfrac{(-1)^{i-1}}{i^p} = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q}}{(1-x_1 x_2 \cdots x_q)(1+x_1 x_2 \cdots x_{p+q})}$$ $$D(p,q) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k^q} \sum_{i=1}^k \dfrac{(-1)^{i-1}}{i^p} = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q}}{(1+x_1 x_2 \cdots x_q)(1-x_1 x_2 \cdots x_{p+q})}$$

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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, $$\sum_{k=1}^{\infty} \dfrac{1}{k^q} \sum_{n=1}^k \dfrac{1}{n^p} \sum_{m=1}^n \dfrac{1}{m^r} = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q+r \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q+r}}{(1- x_1 \cdots x_q)(1-x_1 \cdots x_{p+q}) \cdots (1-x_1 \cdots x_{p+q+r})}$$ $$\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k^q} \sum_{n=1}^k \dfrac{1}{n^p} \sum_{m=1}^n \dfrac{1}{m^r} = \underbrace{\int_0^1 \cdots \int_0^1}_{p+q+r \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q+r}}{(1+ x_1 \cdots x_q)(1+x_1 \cdots x_{p+q}) \cdots (1+x_1 \cdots x_{p+q+r})}$$ $$\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k^q} \sum_{n=1}^k \dfrac{(-1)^{n-1}}{n^p} \sum_{m=1}^n \dfrac{1}{m^r} = \underbrace{\int_0^1 \cdots \int_0^1}_{p+q+r \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q+r}}{(1+ x_1 \cdots x_q)(1-x_1 \cdots x_{p+q}) \cdots (1-x_1 \cdots x_{p+q+r})}$$ $$\sum_{k=1}^{\infty} \dfrac{1}{k^q} \sum_{n=1}^k \dfrac{(-1)^{n-1}}{n^p} \sum_{m=1}^n \dfrac{1}{m^r} = \underbrace{\int_0^1 \cdots \int_0^1}_{p+q+r \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q+r}}{(1- x_1 \cdots x_q)(1+x_1 \cdots x_{p+q}) \cdots (1+x_1 \cdots x_{p+q+r})}$$

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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.

Answer 2

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$$ A(1,1) = \frac{1}{2} \left( \zeta(2) - \log^2(2)\right) $$

Answer 3

Related problems: (I), (II), (III), (IV), $(5)$. For $A(1, 1)$, one can have the integral representation

$$ A(1,1) = \int _{1}^{2}\!{\frac {\ln \left( t \right) }{t \left( t-1 \right) }} {dw}.$$

In general, one can have the following representation for $A(p,1)$

$$ A(p,1) = -\int _{0}^{1}\!{\frac { Li_{p}\left( -u \right) }{ \left( 1+ u \right) u}}{du},$$

where $Li_{p}(-u)$ is the polylogarithm function. Here are some numerical values for $p$ from $1$ to $5$

$$ 0.5822405265,\, 0.6319661978,\, 0.6603570751,\, 0.6759332433,\, 0.6842426955. $$

The General Case A(p,q):

$$ A(p,q) =\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q} = \frac{\left( -1 \right) ^{q}}{\Gamma(q)}\int _{0}^{1}\!{\frac { \left( \ln\left( u \right) \right)^{q-1}{Li_{p}(-u)} }{ u\left( 1+ u \right) }}{du}. $$

Some numerical values

$$ A(1,2) = .7512855645,\, A(2, 3) = .8793713030, \, A(3, 4) = .9407280160, $$

$$ A(2,1) = .6319661978, A(3, 2) = .8024944234, A(4, 3) = .8956823180. $$

Added

The General Case B(p,q):

$$ B(p,q) = \sum_{k=1}^{\infty} \dfrac{H_k^{(p)}}{k^q}=\frac{(-1)^q}{\Gamma(q)}\int_{0}^{1}\!{\frac {\left(\ln\left(u\right)\right)^{q-1}{Li_{p}(u)} }{ u\left( u-1 \right)}}{du}. $$

Some numerical values

$$ B(1, 2) = 2.404113806, B(2, 3) = 1.265738152, B(3, 4) = 1.093509100, $$

$$ B(3, 2) = 1.748493953, B(4, 3) = 1.215854292, B(5, 4) = 1.084986223. $$

Answer 4

Let us start by noting that the first two sums below are the same (interchange the summation variables and the order of the sums) \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{n(n+m)} + \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{m(n+m)} = \left( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \right) \left( \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m} \right). \end{eqnarray*} Thus, we have \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{n(n+m)} = \frac{(\ln(2))^2}{2}. \end{eqnarray*} Now \begin{eqnarray*} A(1,1) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} &=& \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2} - \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{n(n+m)} \\ &=& \frac{1}{2} \zeta_2 - \frac{1}{2} ( \ln(2) )^2. \end{eqnarray*}

Consider the Harmonic numbers in two ways \begin{eqnarray*} H_n=\sum_{k=1}^{n} \frac{1}{k} = \sum_{m=1}^{\infty} \left( \frac{1}{m} -\frac{1}{m+n} \right). \end{eqnarray*} We have \begin{eqnarray*} \sum_{k=1}^{\infty} \frac{H_k}{k^2} &=& \sum_{m=1}^{\infty} \frac{1}{m^3} + \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n(n+m)^2} \\ &=& \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+m)} . \end{eqnarray*} As we saw earlier, the first two sums below are the same (interchange the summation variables and the order of the sums) \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n(n+m)^2} + \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m(n+m)^2} = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+m)}. \end{eqnarray*} After a little bit of algebra \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+m)} = 2 \zeta_3 \\ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n(n+m)^2} = \zeta_3 . \\ \end{eqnarray*} Next, split the sum $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+m)}$ according to weather $m>n,m=n$ and $m<n$, this gives \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+m)} = 2 \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n(n+m)(2n+m)} +\frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n^3} \end{eqnarray*} So \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n(n+m)(2n+m)} = \frac{3}{4} \zeta_3. \end{eqnarray*} Partial fractions ... \begin{eqnarray*} \underbrace{\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+m)}}_{2 \zeta_3} + \underbrace{\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n(n+m)(2n+m)}}_{\frac{3}{4} \zeta_3} = 2 \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(2n+m)} \end{eqnarray*} gives \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(2n+m)} = \frac{11}{8} \zeta_3. \end{eqnarray*} Partial fractions ... \begin{eqnarray*} \underbrace{\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(2n+m)}}_{\frac{11}{8} \zeta_3} + \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m(n+m)(2n+m)} = \underbrace{\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+m)}}_{2 \zeta_3} \end{eqnarray*} gives \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m(n+m)(2n+m)} = \frac{5}{8} \zeta_3. \end{eqnarray*} Next, consider the sum $ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n(n+m)(2n+m)} $ according to weather $m$ is odd or even \begin{eqnarray*} \underbrace{\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n(n+m)(2n+m)}}_{ \frac{3}{4} \zeta_3} = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n(2m-1)(2n+2m-1)}+ \frac{1}{2} \underbrace{ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n(n+m)(n+2m)}}_{ \frac{5}{8} \zeta_3} \end{eqnarray*} so \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n(2m-1)(2n+2m-1)}= \frac{7}{16} \zeta_3. \end{eqnarray*} Again consider weather $m$ is odd or even \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{n(n+m)(2n+m)} = \underbrace{\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n(2m-1)(2n+2m-1)}}_{ \frac{7}{16} \zeta_3} - \frac{1}{2} \underbrace{ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n(n+m)(n+2m)}}_{ \frac{5}{8} \zeta_3} \end{eqnarray*} so \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{n(n+m)(2n+m)}= \frac{1}{8} \zeta_3. \end{eqnarray*} Now, split the sum $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{m+n}}{nm(n+m)}$ according to weather $m>n,m=n$ and $m<n$, \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{m+n}}{nm(n+m)} = -2 \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{n(n+m)(2n+m)} +\frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n^3} \end{eqnarray*} So \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{m+n}}{nm(n+m)} = \frac{1}{4} \zeta_3. \end{eqnarray*} Again, the first two sums below are equal \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{n(n+m)^2} + \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{m(n+m)^2} = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{nm(n+m)} \end{eqnarray*} so \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{n(n+m)^2} = \frac{1}{8} \zeta_3. \end{eqnarray*} Note that \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{n^2(n+m)} + \underbrace{\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{nm(n+m)}}_{\frac{1}{4} \zeta_3} = \underbrace{\left( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} \right) }_{\frac{1}{2} \zeta_2 } \underbrace{\left( \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m} \right)}_{\ln(2)}. \end{eqnarray*} Thus, we have \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{n^2(n+m)} = \frac{1}{2} \zeta_2 \ln(2) - \frac{1}{4} \zeta_3. \end{eqnarray*}

So ... finally ... \begin{eqnarray*} A(1,2) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k^2} &=& \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^3} - \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{n(n+m)^2} \\ &=& \frac{5}{8} \zeta_3 \end{eqnarray*} and \begin{eqnarray*} A(2,1) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k^{(2)}}{k} &=& \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^3} - \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{n^2(n+m)} \\ &=& \frac{1}{2} \zeta_3 - \frac{1}{2} \zeta_2 \ln(2). \end{eqnarray*}

Answer 5

$$\begin{array}{l}{I_1} = \int\limits_0^1 {\frac{{x\ln \left( {1 - x} \right)}}{{1 + {x^2}}}\,dx\,\,\,,\,\,\,{I_2} = } \int\limits_0^1 {\frac{{x\ln \left( {1 + x} \right)}}{{1 + {x^2}}}\,dx\,\,\,} \\{I_1}\left( m \right) = \frac{1}{2}\int\limits_0^1 {\frac{{2x{{\left( {1 - x} \right)}^m}}}{{1 + {x^2}}}dx\,\,\,\,,\,{I_1}^{\rm{\backslash }}\left( {m = 0} \right) = I\,\,\,\,\,,{I_1}\left( m \right) = \left( {\frac{1}{2}{{\left( {1 - x} \right)}^m}\ln \left( {1 + {x^2}} \right)} \right)} _0^1 + \frac{m}{2}\int\limits_0^1 {\ln \left( {1 + {x^2}} \right)} {\left( {1 - x} \right)^{m - 1}}dx\\ = - \frac{m}{2}\sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{k}\int\limits_0^1 {{x^{2k}}{{\left( {1 - x} \right)}^{m - 1}}dx = } } - \frac{m}{2}\sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{k}\left( {\frac{{\Gamma \left( {2k + 1} \right)\Gamma \left( m \right)}}{{\Gamma \left( {2k + m + 1} \right)}}} \right)} \\\frac{{d{I_1}\left( m \right)}}{{dm}} = - \frac{1}{2}\sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^k}\Gamma \left( {2k + 1} \right)}}{k}\left( {\frac{{\Gamma \left( {2k + m + 1} \right){\Gamma ^{\rm{\backslash }}}\left( {m + 1} \right) - \Gamma \left( {m + 1} \right){\Gamma ^{\rm{\backslash }}}\left( {2k + m + 1} \right)}}{{{\Gamma ^2}\left( {2k + m + 1} \right)}}} \right)} \end{array} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGjb % WaaSbaaSqaaiaaigdaaeqaaOGaeyypa0Zaa8qCaeaadaWcaaqaaiaa % dIhaciGGSbGaaiOBamaabmaabaGaaGymaiabgkHiTiaadIhaaiaawI % cacaGLPaaaaeaacaaIXaGaey4kaSIaamiEamaaCaaaleqabaGaaGOm % aaaaaaGccaaMc8UaamizaiaadIhacaaMc8UaaGPaVlaaykW7caGGSa % GaaGPaVlaaykW7caaMc8UaamysamaaBaaaleaacaaIYaaabeaakiab % g2da9aWcbaGaaGimaaqaaiaaigdaa0Gaey4kIipakmaapehabaWaaS % aaaeaacaWG4bGaciiBaiaac6gadaqadaqaaiaaigdacqGHRaWkcaWG % 4baacaGLOaGaayzkaaaabaGaaGymaiabgUcaRiaadIhadaahaaWcbe % qaaiaaikdaaaaaaOGaaGPaVlaadsgacaWG4bGaaGPaVlaaykW7caaM % c8oaleaacaaIWaaabaGaaGymaaqdcqGHRiI8aaGcbaGaamysamaaBa % aaleaacaaIXaaabeaakmaabmaabaGaamyBaaGaayjkaiaawMcaaiab % g2da9maalaaabaGaaGymaaqaaiaaikdaaaWaa8qCaeaadaWcaaqaai % aaikdacaWG4bWaaeWaaeaacaaIXaGaeyOeI0IaamiEaaGaayjkaiaa % wMcaamaaCaaaleqabaGaamyBaaaaaOqaaiaaigdacqGHRaWkcaWG4b % WaaWbaaSqabeaacaaIYaaaaaaakiaadsgacaWG4bGaaGPaVlaaykW7 % caaMc8UaaGPaVlaacYcacaaMc8UaamysamaaBaaaleaacaaIXaaabe % aakmaaCaaaleqabaGaafixaaaakmaabmaabaGaamyBaiabg2da9iaa % icdaaiaawIcacaGLPaaacqGH9aqpcaWGjbGaaGPaVlaaykW7caaMc8 % UaaGPaVlaaykW7caGGSaGaamysamaaBaaaleaacaaIXaaabeaakmaa % bmaabaGaamyBaaGaayjkaiaawMcaaiabg2da9maabmaabaWaaSaaae % aacaaIXaaabaGaaGOmaaaadaqadaqaaiaaigdacqGHsislcaWG4baa % caGLOaGaayzkaaWaaWbaaSqabeaacaWGTbaaaOGaciiBaiaac6gada % qadaqaaiaaigdacqGHRaWkcaWG4bWaaWbaaSqabeaacaaIYaaaaaGc % caGLOaGaayzkaaaacaGLOaGaayzkaaaaleaacaaIWaaabaGaaGymaa % qdcqGHRiI8aOWaa0baaSqaaiaaicdaaeaacaaIXaaaaOGaey4kaSYa % aSaaaeaacaWGTbaabaGaaGOmaaaadaWdXbqaaiGacYgacaGGUbWaae % WaaeaacaaIXaGaey4kaSIaamiEamaaCaaaleqabaGaaGOmaaaaaOGa % ayjkaiaawMcaaaWcbaGaaGimaaqaaiaaigdaa0Gaey4kIipakmaabm % aabaGaaGymaiabgkHiTiaadIhaaiaawIcacaGLPaaadaahaaWcbeqa % aiaad2gacqGHsislcaaIXaaaaOGaamizaiaadIhaaeaacqGH9aqpcq % GHsisldaWcaaqaaiaad2gaaeaacaaIYaaaamaaqahabaWaaSaaaeaa % daqadaqaaiabgkHiTiaaigdaaiaawIcacaGLPaaadaahaaWcbeqaai % aadUgaaaaakeaacaWGRbaaamaapehabaGaamiEamaaCaaaleqabaGa % aGOmaiaadUgaaaGcdaqadaqaaiaaigdacqGHsislcaWG4baacaGLOa % GaayzkaaWaaWbaaSqabeaacaWGTbGaeyOeI0IaaGymaaaakiaadsga % caWG4bGaeyypa0daleaacaaIWaaabaGaaGymaaqdcqGHRiI8aaWcba % Gaam4Aaiabg2da9iaaigdaaeaacqGHEisPa0GaeyyeIuoakiabgkHi % TmaalaaabaGaamyBaaqaaiaaikdaaaWaaabCaeaadaWcaaqaamaabm % aabaGaeyOeI0IaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaam4A % aaaaaOqaaiaadUgaaaWaaeWaaeaadaWcaaqaaiabfo5ahnaabmaaba % GaaGOmaiaadUgacqGHRaWkcaaIXaaacaGLOaGaayzkaaGaeu4KdC0a % aeWaaeaacaWGTbaacaGLOaGaayzkaaaabaGaeu4KdC0aaeWaaeaaca % aIYaGaam4AaiabgUcaRiaad2gacqGHRaWkcaaIXaaacaGLOaGaayzk % aaaaaaGaayjkaiaawMcaaaWcbaGaam4Aaiabg2da9iaaigdaaeaacq % GHEisPa0GaeyyeIuoaaOqaamaalaaabaGaamizaiaadMeadaWgaaWc % baGaaGymaaqabaGcdaqadaqaaiaad2gaaiaawIcacaGLPaaaaeaaca % WGKbGaamyBaaaacqGH9aqpcqGHsisldaWcaaqaaiaaigdaaeaacaaI % YaaaamaaqahabaWaaSaaaeaadaqadaqaaiabgkHiTiaaigdaaiaawI % cacaGLPaaadaahaaWcbeqaaiaadUgaaaGccqqHtoWrdaqadaqaaiaa % ikdacaWGRbGaey4kaSIaaGymaaGaayjkaiaawMcaaaqaaiaadUgaaa % WaaeWaaeaadaWcaaqaaiabfo5ahnaabmaabaGaaGOmaiaadUgacqGH % RaWkcaWGTbGaey4kaSIaaGymaaGaayjkaiaawMcaaiabfo5ahnaaCa % aaleqabaGaagixaaaakmaabmaabaGaamyBaiabgUcaRiaaigdaaiaa % wIcacaGLPaaacqGHsislcqqHtoWrdaqadaqaaiaad2gacqGHRaWkca % aIXaaacaGLOaGaayzkaaGaeu4KdC0aaWbaaSqabeaacayGCbaaaOWa % aeWaaeaacaaIYaGaam4AaiabgUcaRiaad2gacqGHRaWkcaaIXaaaca % GLOaGaayzkaaaabaGaeu4KdC0aaWbaaSqabeaacaaIYaaaaOWaaeWa % aeaacaaIYaGaam4AaiabgUcaRiaad2gacqGHRaWkcaaIXaaacaGLOa % GaayzkaaaaaaGaayjkaiaawMcaaaWcbaGaam4Aaiabg2da9iaaigda % aeaacqGHEisPa0GaeyyeIuoaaaaa!54A6! $$ $$\begin{array}{l}{I_1} = \frac{1}{2}\sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{k}} \left( {\gamma + \psi \left( {2k + 1} \right)} \right) = \frac{1}{2}\sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{k}} {H_{2k}}\\from\,\,\int\limits_0^1 {{x^{k - 1}}\ln \left( {1 - x} \right)\,dx = - \frac{{{H_k}}}{k}} \\\sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{2k}}} {H_{2k}} - \sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{2k}}} {H_k} = \sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^k}} \int\limits_0^1 {{x^{2k - 1}}\ln \left( {1 + x} \right)\,dx} = \int\limits_0^1 {\frac{{x\ln \left( {1 + x} \right)}}{{1 + {x^2}}}\,dx\, = {I_2}\,\,} \\{I_1} + {I_2} = \sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{2k}}} {H_k} \Rightarrow \left( 1 \right)\end{array} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGjb % WaaSbaaSqaaiaaigdaaeqaaOGaeyypa0ZaaSaaaeaacaaIXaaabaGa % aGOmaaaadaaeWbqaamaalaaabaWaaeWaaeaacqGHsislcaaIXaaaca % GLOaGaayzkaaWaaWbaaSqabeaacaWGRbaaaaGcbaGaam4AaaaaaSqa % aiaadUgacqGH9aqpcaaIXaaabaGaeyOhIukaniabggHiLdGcdaqada % qaaiabeo7aNjabgUcaRiabeI8a5naabmaabaGaaGOmaiaadUgacqGH % RaWkcaaIXaaacaGLOaGaayzkaaaacaGLOaGaayzkaaGaeyypa0ZaaS % aaaeaacaaIXaaabaGaaGOmaaaadaaeWbqaamaalaaabaWaaeWaaeaa % cqGHsislcaaIXaaacaGLOaGaayzkaaWaaWbaaSqabeaacaWGRbaaaa % GcbaGaam4AaaaaaSqaaiaadUgacqGH9aqpcaaIXaaabaGaeyOhIuka % niabggHiLdGccaWGibWaaSbaaSqaaiaaikdacaWGRbaabeaaaOqaai % aadAgacaWGYbGaam4Baiaad2gacaaMc8UaaGPaVpaapehabaGaamiE % amaaCaaaleqabaGaam4AaiabgkHiTiaaigdaaaGcciGGSbGaaiOBam % aabmaabaGaaGymaiabgkHiTiaadIhaaiaawIcacaGLPaaacaaMc8Ua % amizaiaadIhacqGH9aqpcqGHsisldaWcaaqaaiaadIeadaWgaaWcba % Gaam4AaaqabaaakeaacaWGRbaaaaWcbaGaaGimaaqaaiaaigdaa0Ga % ey4kIipaaOqaamaaqahabaWaaSaaaeaadaqadaqaaiabgkHiTiaaig % daaiaawIcacaGLPaaadaahaaWcbeqaaiaadUgaaaaakeaacaaIYaGa % am4AaaaaaSqaaiaadUgacqGH9aqpcaaIXaaabaGaeyOhIukaniabgg % HiLdGccaWGibWaaSbaaSqaaiaaikdacaWGRbaabeaakiabgkHiTmaa % qahabaWaaSaaaeaadaqadaqaaiabgkHiTiaaigdaaiaawIcacaGLPa % aadaahaaWcbeqaaiaadUgaaaaakeaacaaIYaGaam4AaaaaaSqaaiaa % dUgacqGH9aqpcaaIXaaabaGaeyOhIukaniabggHiLdGccaWGibWaaS % baaSqaaiaadUgaaeqaaOGaeyypa0ZaaabCaeaadaqadaqaaiabgkHi % TiaaigdaaiaawIcacaGLPaaadaahaaWcbeqaaiaadUgaaaaabaGaam % 4Aaiabg2da9iaaigdaaeaacqGHEisPa0GaeyyeIuoakmaapehabaGa % aiiEamaaCaaaleqabaGaaGOmaiaadUgacqGHsislcaaIXaaaaOGaci % iBaiaac6gadaqadaqaaiaaigdacqGHRaWkcaWG4baacaGLOaGaayzk % aaGaaGPaVlaadsgacaWG4baaleaacaaIWaaabaGaaGymaaqdcqGHRi % I8aOGaeyypa0Zaa8qCaeaadaWcaaqaaiaadIhaciGGSbGaaiOBamaa % bmaabaGaaGymaiabgUcaRiaadIhaaiaawIcacaGLPaaaaeaacaaIXa % Gaey4kaSIaamiEamaaCaaaleqabaGaaGOmaaaaaaGccaaMc8Uaamiz % aiaadIhacaaMc8Uaeyypa0JaamysamaaBaaaleaacaaIYaaabeaaki % aaykW7caaMc8oaleaacaaIWaaabaGaaGymaaqdcqGHRiI8aaGcbaGa % amysamaaBaaaleaacaaIXaaabeaakiabgUcaRiaadMeadaWgaaWcba % GaaGOmaaqabaGccqGH9aqpdaaeWbqaamaalaaabaWaaeWaaeaacqGH % sislcaaIXaaacaGLOaGaayzkaaWaaWbaaSqabeaacaWGRbaaaaGcba % GaaGOmaiaadUgaaaaaleaacaWGRbGaeyypa0JaaGymaaqaaiabg6Hi % LcqdcqGHris5aOGaamisamaaBaaaleaacaWGRbaabeaakiabgkDiEp % aabmaabaGaaGymaaGaayjkaiaawMcaaaaaaa!EDD5! $$ $$\begin{array}{l}{I_1} + {I_2} = \int\limits_0^1 {\frac{{x\ln \left( {1 - {x^2}} \right)}}{{1 + {x^2}}}\,dx\,\,\,} = \frac{1}{2}\int\limits_0^1 {\frac{{\ln \left( {1 - x} \right)}}{{1 + x}}\,dx\,\,\,} \\by\,\,\,using\,\,the\,\,x = \frac{{1 - y}}{{1 + y}}\,\,\,\,\\\sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{2k}}} {H_k} = {I_1} + {I_2} = \frac{1}{2}\int\limits_0^1 {\frac{{\ln \left( {\frac{{2y}}{{1 + y}}} \right)}}{{1 + y}}\,dy = \frac{1}{4}{{\ln }^2}2 - \frac{{{\pi ^2}}}{{24}}} \\and\,\,then\,\,we\,\,obtain\,\,the\,\,needed\,\,proof\end{array} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGjb % WaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaamysamaaBaaaleaacaaI % Yaaabeaakiabg2da9maapehabaWaaSaaaeaacaWG4bGaciiBaiaac6 % gadaqadaqaaiaaigdacqGHsislcaWG4bWaaWbaaSqabeaacaaIYaaa % aaGccaGLOaGaayzkaaaabaGaaGymaiabgUcaRiaadIhadaahaaWcbe % qaaiaaikdaaaaaaOGaaGPaVlaadsgacaWG4bGaaGPaVlaaykW7caaM % c8oaleaacaaIWaaabaGaaGymaaqdcqGHRiI8aOGaeyypa0ZaaSaaae % aacaaIXaaabaGaaGOmaaaadaWdXbqaamaalaaabaGaciiBaiaac6ga % daqadaqaaiaaigdacqGHsislcaWG4baacaGLOaGaayzkaaaabaGaaG % ymaiabgUcaRiaadIhaaaGaaGPaVlaadsgacaWG4bGaaGPaVlaaykW7 % caaMc8oaleaacaaIWaaabaGaaGymaaqdcqGHRiI8aaGcbaGaamOyai % aadMhacaaMc8UaaGPaVlaaykW7caWG1bGaam4CaiaadMgacaWGUbGa % am4zaiaaykW7caaMc8UaamiDaiaadIgacaWGLbGaaGPaVlaaykW7ca % WG4bGaeyypa0ZaaSaaaeaacaaIXaGaeyOeI0IaamyEaaqaaiaaigda % cqGHRaWkcaWG5baaaiaaykW7caaMc8UaaGPaVlaaykW7aeaadaaeWb % qaamaalaaabaWaaeWaaeaacqGHsislcaaIXaaacaGLOaGaayzkaaWa % aWbaaSqabeaacaWGRbaaaaGcbaGaaGOmaiaadUgaaaaaleaacaWGRb % Gaeyypa0JaaGymaaqaaiabg6HiLcqdcqGHris5aOGaamisamaaBaaa % leaacaWGRbaabeaakiabg2da9iaadMeadaWgaaWcbaGaaGymaaqaba % GccqGHRaWkcaWGjbWaaSbaaSqaaiaaikdaaeqaaOGaeyypa0ZaaSaa % aeaacaaIXaaabaGaaGOmaaaadaWdXbqaamaalaaabaGaciiBaiaac6 % gadaqadaqaamaalaaabaGaaGOmaiaadMhaaeaacaaIXaGaey4kaSIa % amyEaaaaaiaawIcacaGLPaaaaeaacaaIXaGaey4kaSIaamyEaaaaca % aMc8UaamizaiaadMhacqGH9aqpdaWcaaqaaiaaigdaaeaacaaI0aaa % aiGacYgacaGGUbWaaWbaaSqabeaacaaIYaaaaOGaaGOmaiabgkHiTm % aalaaabaGaeqiWda3aaWbaaSqabeaacaaIYaaaaaGcbaGaaGOmaiaa % isdaaaaaleaacaaIWaaabaGaaGymaaqdcqGHRiI8aaGcbaGaamyyai % aad6gacaWGKbGaaGPaVlaaykW7caWG0bGaamiAaiaadwgacaWGUbGa % aGPaVlaaykW7caWG3bGaamyzaiaaykW7caaMc8Uaam4Baiaadkgaca % WG0bGaamyyaiaadMgacaWGUbGaaGPaVlaaykW7caWG0bGaamiAaiaa % dwgacaaMc8UaaGPaVlaad6gacaWGLbGaamyzaiaadsgacaWGLbGaam % izaiaaykW7caaMc8UaamiCaiaadkhacaWGVbGaam4BaiaadAgaaaaa % !ED28! $$ by Mr. Sherif Hamed

Answer 6

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