# 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.

-
Write out the infinite product $\prod_{k=1}^\infty(1-x^k/k^s)$ in terms of the gamma function, use veitas formulas to gather up partial sums of 1/k^s, and align them with the products corresponding power series, euler did something similar to solve the basel problem. Try doing it for the case s=2, in which case the product is sin(pix)/(pix). –  Ethan Jan 11 '13 at 5:34
@Ethan: Spell out the details, and perhaps that becomes an answer? :) –  Mike Spivey Jan 11 '13 at 5:36
Can you inculde the defintion of $A(p,q)$ ? –  Amr Jan 11 '13 at 6:07
@Ethan I missed that –  Amr Jan 11 '13 at 6:08
It is interesting that this question has been marked as a favorite $6$ favorites whereas it only has $4$ up-votes. –  user17762 Jan 11 '13 at 22:08

$A(1,1)$: \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 $$\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}H_n=\color{#00A000}{\frac12\zeta(2)}-\color{#0000FF}{\frac12\log(2)^2}\tag{3}$$

-
Very nice answer! I hope it will be highly upvoted! –  Chris's sis Sep 20 '13 at 15:48
@Chris'ssis: thanks. I'm not holding my breath since I came across this question so late. –  robjohn Sep 20 '13 at 15:50
very good answer (+1) . –  what'sup Sep 20 '13 at 18:18
Rob, this is very nice! I particularly like the fact that it is the kind of approach I was originally trying but couldn't get to work. For others reading this, what Rob has done is to extract the eta function partial sum and then show that the rest turns into the square of the partial sum of the alternating harmonic series plus change. It's really clear how the answer falls out this way. Beautiful. –  Mike Spivey Sep 20 '13 at 21:29
@MikeSpivey: Thanks for the comments and corrections! Let me know if you find anything that needs explanation for my derivation of $A(1,2)$. –  robjohn Sep 20 '13 at 21:43

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

$$\sum_{k\geq 1}\frac{H^{(p)}_k}{k}x^k=\mathrm{Li}_{p+1}(x)+\int^x_0 \frac{\mathrm{Li}_p(t)}{1-t}\,d t$$

$$\sum_{k\geq 1}\frac{H_k}{k}x^k=\mathrm{Li}_{2}(x)+\frac{1}{2}\log^2(1-x)$$

$$\sum_{k\geq 1}\frac{H_k}{k}(-1)^k=-\frac{\pi^2}{12}+\frac{1}{2}\log^2(2)$$

-
Typo on the last line. Should be $\sum_{k\geq 1}\frac{H_k}{k}(-1)^k$, I believe. –  columbus8myhw Aug 24 at 2:45
@columbus8myhw, thanks. –  Zaid Alyafeai 10 hours ago

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

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

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

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

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

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

-
Thanks, but that's already in part of Marvis's answer. –  Mike Spivey Jan 12 '13 at 23:04
@MikeSpivey: You are welcome. See the more general case $A(p,1)$. –  Mhenni Benghorbal Jan 12 '13 at 23:35
Your generalization to the $A(p,1)$ case is interesting. Do you have a proof or reference for that? –  Mike Spivey Jan 13 '13 at 3:09
A proof for $A(p,1)$ is as follows. $$A(p,1) = \underbrace{\int_0^1 \cdots \int_0^1}_{p+1 \text{ times}} \int_0^1 \dfrac{dx_1 dx_2 \cdots dx_{p+1}}{(1+x_1)(1+x_1x_2x_3 \cdots x_{p+1})}$$ Now $$\underbrace{\int_0^1 \cdots \int_0^1}_{p \text{ times}} \int_0^1 \dfrac{dx_2 dx_3 \cdots dx_{p+1}}{(1+x_1x_2x_3 \cdots x_{p})} = -\dfrac{\text{Li}_p(-x_1)}{x_1}$$ which follows immediately from induction and definition of $\text{Li}_n(x)$. –  user17762 Jan 13 '13 at 3:20
@Marvis: Thanks. +1 to Mhenni's answer. –  Mike Spivey Jan 13 '13 at 3:28


\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}

-
How do you go from $\int_0^1 \frac{1-t^k}{1-t} \, dt$ to $\int_0^1 \ln(1-t) (-kt^{k-1}) \, dt$ in the second step? –  Mike Spivey Jun 9 at 18:43
@MikeSpivey Integrating by parts since $\displaystyle\large{{\rm d}\ln\left(1 - t\right) \over {\rm d}t} = -\,{1 \over 1 - t}$. Thanks. –  Felix Marin Jun 9 at 18:46
Got it; thanks! And +1. –  Mike Spivey Jun 9 at 18:49

For $A(1,q)$: note $$H_n=\int_0^1\frac{1-x^n}{1-x}dx, \frac{(-1)^{q-1}(q-1)!}{n^q}=\int_0^1x^{n-1}\ln^{q-1}xdx.$$ So \begin{eqnarray} \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^q}H_n&=&\frac{(-1)^{q-1}}{(q-1)!}\sum_{n=1}^\infty\int_0^1\int_0^1\frac{(-1)^{n-1}(1-x^n)y^{n-1}}{1-x}\ln^{q-1}ydxdy\\ &=&\frac{(-1)^{q-1}}{(q-1)!}\int_0^1\int_0^1\frac{\ln^{q-1}y}{(1+y)(1+xy)}dxdy. \end{eqnarray} So \begin{eqnarray} A(1,1)&=&\int_0^1\int_0^1\frac{1}{(1+y)(1+xy)}dxdy=-\int_0^1\frac{\ln(\frac{1+x}{2})}{1-x}dx=\frac{1}{2}(\zeta(2)-\ln^22),\\ A(1,2)&=&-\int_0^1\int_0^1\frac{\ln y}{(1+y)(1+xy)}dxdy=-\int_0^1\frac{\ln y\ln(1+y)}{y(1+y)}dy=\frac{5\zeta(3)}{8},\\ A(1,3)&=&-\int_0^1\int_0^1\frac{\ln^2 y}{(1+y)(1+xy)}dxdy=\frac{1}{2}\int_0^1\frac{\ln^2 y\ln(1+y)}{y(1+y)}dy=\cdots. \end{eqnarray}

-

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}

-
That is interesting. Thanks for the observation. –  Mike Spivey Jul 24 at 21:40

$A(2,1)$:

\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 $$A(2,1)=\sum_{n=1}^\infty\frac{H_n}{2^nn^2}\tag{8}$$

-
Thanks, Rob. I appreciate you taking the time to add this answer. –  Mike Spivey Dec 17 '13 at 19:34

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.

-
very nice lol, much better then my idea –  Ethan Jan 11 '13 at 6:10
@Marvis: a powerful answer. We need such answers on MSE. (+1) –  Chris's sis Jan 11 '13 at 7:21
I am completely astounded by this brilliant answer! I also succeeded in calculating those series by using dilogarithm identities and trilogarithm identity, my solutions lack a systematic approach that your illuminating answer shows. That's why I love this answer. –  sos440 Jan 11 '13 at 18:11
I feel completely overwhelmed solely by skimming this answer, Marvis. Dare I call it beautiful, for I know not what it is . . . –  000 Jan 13 '13 at 3:22
My answer seems like a dinghy next to your aircraft carrier :-) –  robjohn Sep 20 '13 at 16:04

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 $A(1,1)$

-

$A(1,2)$: \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, $$\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}H_n=\frac58\zeta(3)\tag{7}$$

-
An amazing answer! (+1) If I could I'd upvote it $1000$ times. –  Chris's sis Sep 20 '13 at 16:37
@Chris'ssis: I spent pretty much all day yesterday on this sum for a question you asked in chat, then saw that it was asked in this question, too! :-) –  robjohn Sep 20 '13 at 16:42
an amazing answer (+1) –  what'sup Sep 20 '13 at 18:19
Again, very nice! It would never have occurred to me to use the property that $1 + (-1)^k + (-1)^n + (-1)^{n+k} = 4$ when $n$ and $k$ are even and vanishes otherwise. Well done. –  Mike Spivey Sep 20 '13 at 21:53
@Jeff: See this answer with $q=2$, then note that $$\sum_{n\text { odd}}a_n=\frac12\left(\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty(-1)^{n-1}a_n\right)$$ –  robjohn Mar 1 at 20:33

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

-
Thanks, Sasha. One question, though: I don't follow the step $-\sum_{n=1}^\infty x^n \sum_{k=0}^{n-1} (-1)^k \frac{(-1)^{n-k}}{n-k} = -\sum_{n=0}^\infty (-x)^n \cdot \sum_{k=1}^\infty \frac{(-x)^k}{k}$. –  Mike Spivey Jan 11 '13 at 17:47
Never mind; I got it. You're using the product formula for power series. Thanks again. –  Mike Spivey Jan 11 '13 at 17:50