# Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$

How to analytically prove

$$\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3)$$

where $$H_k = \sum_{n\geq 1}^{k}\frac{1}{n}.$$

So far I developed the following

$$\sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)-\, \text{Li}_3(1-x)+\, \log(1-x) \text{Li}_2(1-x) +\frac{1}{2}\log(x) \log^2(1-x)+\zeta(3)$$

where $\text{Li}_3(x)$ is the trilogarithm .

For the derivation see http://www.mathhelpboards.com/f10/interesting-logarithm-integral-5301/

Update

A frined on another site gave the following answer

• Your problem is a special case of one of these general cases. – Mhenni Benghorbal Aug 2 '13 at 9:49
• Most importantly $a_h(1,3)$ and $a_h(2,2)$ are not evaluated in the thread . I think it is helpful ,though . – Zaid Alyafeai Aug 2 '13 at 10:37
• $A(2,2)\sim0.6563115516$. – Mhenni Benghorbal Aug 2 '13 at 11:40
• I think you are evaluating $\sum_{n=1}^{\infty}\frac{H^2(-1)^{n-1}}{n^2}$ which is different than$\sum_{n=1}^{\infty}\frac{H^{(2)}(-1)^{n-1}}{n^2}$ – Zaid Alyafeai Aug 2 '13 at 21:38


The remaining integrals are evaluated by successive integration by parts. Namely,

\begin{align} \int{\ln^{3}\pars{\pm x} \over 1 - x}\,\dd x & = -\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\int\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{\pm x}\,\dd x \\[5mm] & = -\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\,\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} + 6\int\mrm{Li}_{3}'\pars{x}\ln\pars{\pm x}\,\dd x \\[1cm] & = -\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\,\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} + 6\,\mrm{Li}_{3}\pars{x}\ln\pars{\pm x} \\[5mm] & - 6\int\mrm{Li}_{4}'\pars{x}\,\dd x \\[1cm] & = -\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\,\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} + 6\,\mrm{Li}_{3}\pars{x}\ln\pars{\pm x} \\[5mm] & - 6\,\mrm{Li}_{4}\pars{x}\label{2}\tag{2} \end{align}

With \eqref{1} and \eqref{2}: $$\bbox[15px,#ffe,border:1px dotted navy]{\ds{ \sum_{k\ \geq\ 1}{\pars{-1}^{k} \over k^{3}}H_{k} = -\,{11 \over 360}\,\pi^{4} - {1 \over 12}\ln^{2}\pars{2}\pi^{2} + {1 \over 12}\,\ln^{4}\pars{2} + 2\,\mrm{Li}_{4}\pars{1 \over 2} + {7 \over 4}\,\ln\pars{2}\zeta\pars{3}}}$$

• Good job :).... – Zaid Alyafeai Apr 26 '17 at 6:56
• @ZaidAlyafeai Thanks. A long trip after we see the right way. – Felix Marin Apr 26 '17 at 6:57

Let us first recall that harmonic numbers have generating function \begin{align} \sum_{k=1}^{\infty}H_kx^k=-\frac{\ln(1-x)}{1-x}, \end{align} and therefore \begin{align} S=\sum_{k=1}^{\infty}\frac{(-1)^k}{k^3}H_k&=\frac{1}{2}\sum_{k=1}^{\infty}(-1)^kH_k\int_0^{\infty}e^{-kx}x^2dx=\\ &=-\frac{1}{2}\int_0^{\infty}\frac{\ln(1+e^{-x})}{1+e^{-x}}x^2dx. \end{align} Mathematica knows how to evaluate the last integral in terms of zeta values and polylogarithms. Its answer is $$S=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3).$$ It is unlikely that it can be simplified further: Wolfram Alpha proposes alternative expressions for $\mathrm{Li}_{2}\left(\frac12\right)$ and $\mathrm{Li}_{3}\left(\frac12\right)$ in terms of elementary functions and zeta values, but does not suggest anything simpler for $\mathrm{Li}_{4}\left(\frac12\right)$.

• Woow , that something to prove ! Much interestingly is weather we can simplify $\text{Li}_4\left( \frac{1}{2} \right)$ because we already know $\text{Li}_1\left( \frac{1}{2} \right),\text{Li}_2\left( \frac{1}{2} \right),\text{Li}_3\left( \frac{1}{2} \right)$ . – Zaid Alyafeai Aug 1 '13 at 17:37

Related problems: (I), (II), (III). Your sum is a special case of the following general case which I derived an integral representation for it

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

where $\mathrm{Li}_{p}(z)$ is the polylogarithm function. So, letting $p=1$ and $q=3$ in the above formula gives an integral representation for your sum

$$A(1,3) =\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(1)}_k}{k^3} = \frac{\left( -1 \right) ^{3}}{\Gamma(3)}\int _{0}^{1}\!{\frac { \left( \ln\left( u\right) \right)^{3-1}{\mathrm{Li}_{1}(-u)} }{ u (1+u) }}{du}.$$

$$\implies A(1,3) =\frac{1}{2}\int _{0}^{1}\!{\frac { \left( \ln \left( u \right) \right) ^{2} \ln \left( 1+u \right) }{u\left(1+u\right)}}{du} \sim 0.8592471579.$$

See here for related techniques.

Note:

1) $$\mathrm{Li}_{1}(-u)=-\ln(1+u).$$

• Hey @Mehhnni can you evaluate numerically $A(2,2)$ ? – Zaid Alyafeai Aug 2 '13 at 10:40
• @ZaidAlyafeai: You can use the integral representation and then try to use numerical techniques for integrals. Does this what you mean? Let me try to evaluate the integral representation of $A(2,2)$. – Mhenni Benghorbal Aug 2 '13 at 10:43
• @ZaidAlyafeai: By the way, if you are only interested in finding numerical values of these sums, then you can just use Maple or Mathematica and they will give you right away the result. – Mhenni Benghorbal Aug 2 '13 at 11:43
• Just to make sure $A(2,2) = \sum_{k\geq 1}\frac{(-1)^{k+1}H_k^{(2)}}{k^2}$ where we define $H_k^{(2)}= \sum_{n=1}^k \frac{1}{n^2}$ – Zaid Alyafeai Aug 2 '13 at 21:31

\begin{align} S&=\sum_{n=1}^\infty\frac{(-1)^n H_n}{n^3}=\frac12\int_0^1\frac{\ln^2x}{x}\sum_{n=1}^\infty H_n(-x)^n\ dx=-\frac12\underbrace{\int_0^1\frac{\ln^2x\ln(1+x)}{x(1+x)}\ dx}_{x=(1-y)/y}\\ &=\frac12\underbrace{\int_{1/2}^1\frac{\ln^2((1-x)/x)\ln(x)}{1-x}\ dx}_{x=1-y}=\frac12\int_0^{1/2}\frac{\ln^2(x/(1-x))\ln(1-x)}{x}\ dx\\ &=\frac12\left(\int_0^{1/2}\frac{\ln^2x\ln(1-x)}{x}\ dx+\int_0^{1/2}\frac{\ln^3(1-x)}{x}\ dx\right)-\int_0^{1/2}\frac{\ln x\ln^2(1-x)}{x}\ dx\\ &=\frac12\left(I_1+I_2\right)-I_3 \end{align} Applying IBP for the first integral by setting $$dv=\ln^2x/x$$ and $$u=\ln(1-x)$$ and letting $$x=1-y$$ for the second integral, we have:

\begin{align} I_1+I_2&=\frac13\ln^42+\frac13\int_0^{1/2}\frac{\ln^3x}{1-x}\ dx+\int_{1/2}^1\frac{\ln^3x}{1-x}\ dx\\ &=\frac13\ln^42+\int_0^1\frac{\ln^3x}{1-x}\ dx-\frac23\int_0^{1/2}\frac{\ln^3x}{1-x}\ dx\\ &=\frac13\ln^42-6\zeta(4)-\frac23\sum_{n=1}^\infty\int_0^{1/2}x^{n-1}\ln^3x\ dx\\ &=\frac13\ln^42-6\zeta(4)+\frac23\sum_{n=1}^\infty\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}\right)\\ &=4\operatorname{Li_4}\left(\frac12\right)+4\ln2\operatorname{Li_3}\left(\frac12\right)+2\ln^22\operatorname{Li_2}\left(\frac12\right)+\ln^42-6\zeta(4) \end{align} Applying IBP for the third integral by setting $$dv=\ln x/x$$ and $$u=\ln^2(1-x)$$, \begin{align} I_3=\int_0^{1/2}\frac{\ln x\ln^2(1-x)}{x}\ dx&=\frac12\ln^42+\underbrace{\int_0^{1/2}\frac{\ln^2x\ln(1-x)}{1-x}\ dx}_{x=1-y}\\ &=\frac12\ln^42+\int_{1/2}^1\frac{\ln x\ln^2(1-x)}{x}\ dx \end{align} By adding the third integral to both sides, we get: \begin{align} I_3&=\frac14\ln^42+\frac12\int_0^1\frac{\ln x\ln^2(1-x)}{x}\ dx\\ &=\frac14\ln^42+\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1 x^{n-1}\ln x\ dx\\ &=\frac14\ln^42+\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\left(-\frac{1}{n^2}\right)\\ &=\frac14\ln^42+\zeta(4)-\sum_{n=1}^\infty\frac{H_n}{n^3}\\ &=\frac14\ln^42-\frac14\zeta(4) \end{align} Grouping $$I_1, I_2$$ and $$I_3$$: \begin{align} S&=2\operatorname{Li_4}\left(\frac12\right)+2\ln2\operatorname{Li_3}\left(\frac12\right)+\ln^22\operatorname{Li_2}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac14\ln^42\\ &=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42 \end{align} note that we used $$\operatorname{Li_3}\left( \frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$ $$\operatorname{Li_2}\left( \frac12\right) =\frac12\zeta(2)-\frac12\ln^22$$

Bonus:

By Cauchy product we have

$$\operatorname{Li}_2^2(x)=4\sum_{n=1}^\infty x^n\frac{H_n}{n^3}+2\sum_{n=1}^\infty x^n\frac{H_n^{(2)}}{n^2}-6\operatorname{Li}_4(x)$$

set $$x=-1$$ and rearrange the term to have

$$\sum_{n=1}^\infty(-1)^n\frac{H_n^{(2)}}{n^2}=\frac12\operatorname{Li}_2^2(-1)+3\operatorname{Li}_4(-1)-2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^3}$$

substiuting the value of $$\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^3}$$ along with the special values of $$\operatorname{Li}_2(-1)=-\frac12\zeta(2)$$ and $$\operatorname{Li}_4(-1)=-\frac78\zeta(4)$$ we get

$$\sum_{n=1}^{\infty}(-1)^n\frac{H_n^{(2)}}{n^2}=-4\operatorname{Li}_4\left(\frac12\right)+\frac{51}{16}\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac16\ln^42$$

Here we compute the quantity $$A(1,q) := \sum\limits_{k=1}^\infty \frac{(-1)^{k+1} H_k}{k^q} = \frac{(-1)^{q+1}}{(q-1)!} \int\limits_0^1 \frac{[\log(u)]^{q-1} \cdot \log(1+u)}{u(u+1)} d u$$ We have: \begin{eqnarray} A(1,q) &=& \left.\frac{(-1)^{q+1}}{(q-1)!} \frac{\partial^{q-1}}{\partial \theta_1^{q-1}} \frac{\partial^{1}}{\partial \theta_2^{1}} \int\limits_0^1 u^{\theta_1-1} \cdot (u+1)^{\theta_2-1} du \right|_{\theta_1=0,\theta_2=0}\\ &=& \left.\frac{(-1)^{q+1}}{(q-1)!} \frac{\partial^{q-1}}{\partial \theta_1^{q-1}} \frac{\partial^{1}}{\partial \theta_2^{1}} \frac{2^{\theta_2-1}}{\theta_1} F_{2,1}[1, 1-\theta_2,1+\theta_1; 1/2] \right|_{\theta_1=0,\theta_2=0} \\ &=&\left.\frac{(-1)^{q+1}}{(q-1)!} \frac{\partial^{q-1}}{\partial \theta_1^{q-1}} \frac{\log(2) F_{2,1}[1,1,1+\theta_1;1/2] - F_{2,1}^{(0,1,0,0)}[1,1,1+\theta_1;1/2]}{2 \theta_1} \right|_{\theta_1=0} \\ &=& \left.\frac{1}{2} \sum\limits_{l=0}^{q-1} \frac{(-1)^l}{l!} \left[\log(2) \cdot F_{2,1}^{(0,0,l,0)}[1,1,1+\theta_1;1/2] - F_{2,1}^{(0,1,l,0)}[1,1,1+\theta_1;1/2]\right] \frac{1}{\theta_1^{q-l}} \right|_{\theta_1=0} \end{eqnarray} In the second line we integrated by parts and used the definition of the hypergeometric function. In the third line we differentiated with respect to $\theta_2$ and took the limit $\theta_2 \rightarrow 0$ and we defined $$F_{2,1}^{(0,p,q,0)}[a,b,c;x] := \frac{\partial^p}{\partial b^p} \frac{\partial^q}{\partial c^q} F_{2,1}[a,b,c;x]$$ and finally in the last line we applied the chain rule. Now clearly the last expression on the right hand side is singular at zero. Therefore we need to reduce it to the common denominator and then apply the d'Hospital rule. Here the common denominator is $\theta_1^q$ and we have to differentiate $q$-times both the numerator and the denominator. The final result is quite simple. It reads: $$A(1,q) = \frac{(-1)^{q+1}}{2^q q!} \left[ q \log(2) \left( \Psi^{(q-1)}(\frac{1}{2}) - \Psi^{(q-1)}(1) \right) - 2^{q-1} F_{2,1}^{(0,1,q,0)}[1,1,1,;1/2] \right]$$ The last thing that remains to be checked if the higher derivatives of the hipergeometric function all reduce to polygamma function values at unity and at one half. I will check this later.

Now, using the integral representation of the hipergeometric function we fairly easy find the derivative in question. It reads: $$\frac{1}{2} \frac{(-1)^{q-1}}{q!} F_{2,1}^{(0,1,q,0)}[1,1,1;1/2] = \left(1+\log(2)\right) Li_q(-1) - \frac{(-1)^{q-1}}{q!} \int\limits_0^1 \frac{\log(\xi)^q \cdot \log(1+\xi)}{(1+\xi)^2} d\xi$$ Inserting this into the equation for $A(1,q)$ after using Interesting connection between polylogarithms and polygamma functions. we get: $$A(1,q) = -Li_q(-1) + \frac{(-1)^{q-1}}{q!} \int\limits_0^1 \frac{\log(\xi)^q \cdot \log(1+\xi)}{(1+\xi)^2} d\xi$$ and finally using Antiderivative of a function involving logarithms and a fraction. we get the final result for $q=3$. We have: \begin{eqnarray} &&A(1,q) = -Li_q(-1) + \frac{(-1)^{q-1}}{q!} \cdot \\ && \!\!\!\!\!\!\!\left( -6 \text{Li}_4\left(\frac{1}{2}\right)+6 \text{Li}_4(2)-\frac{9 \zeta (3)}{2}-\frac{21}{2} \zeta (3) \log (2)+\frac{\pi ^4}{20}-\frac{\log ^4(2)}{4}+i \pi \log ^3(2)-\frac{1}{2} \pi ^2 \log ^2(2)\right) \\ &&=0.8592471579285906155... \end{eqnarray}

A magical way proposed by Cornel Ioan Valean

Let's consider a powerful form of the Beta function which is presented in the book, (Almost) Impossible Integrals, Sums, and Series, $$\displaystyle \int_0^1 \frac{x^{a-1}+x^{b-1}}{(1+x)^{a+b}} \textrm{d}x = \operatorname{B}(a,b)$$, used for elegant calculations in the Section $$3.7$$, pages $$72$$-$$73$$.

Here is the magic ...

$$\underbrace{\lim_{\substack{a\to0 \\ b \to 0}}\frac{\partial^{3}}{\partial a^2 \partial b}\operatorname{B}(a,b)}_{\displaystyle -5/2\zeta(4)}=3\underbrace{\int_0^1\frac{\log(x)\log^2(1+x)}{x}\textrm{d}x}_{\displaystyle -7/4 \zeta(4)+2 \sum _{n=1}^{\infty} (-1)^{n-1}H_n/n^3}-\underbrace{\int_0^1\frac{\log^2(x)\log(1+x)}{x}\textrm{d}x}_{\displaystyle 7/4\zeta(4)}$$ $$-2\underbrace{\int_0^1 \frac{\log^3(1+x)}{x}\textrm{d}x}_{\displaystyle 6\zeta(4)+3/2\log^2(2)\zeta(2)-21/4\log(2)\zeta(3)\\\displaystyle -\log^4(2)/4-6\operatorname{Li}_4(1/2)},$$

whence we conclude that $$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n}{n^3}$$ $$=\frac{11}{4}\zeta(4)-\frac{7}{4}\log(2)\zeta(3)+\frac{1}{2}\log^2(2)\zeta(2)-\frac{1}{12}\log^4(2)-2\operatorname{Li_4}\left(\frac12\right).$$

Q.E.D.

A first note: Observe the last two integrals in the right-hand side and known and trivial.

A second note: The Beta function limit can be approached in more ways. An elegant way is achieved by Cornel's Master Theorem of Series from the article A master theorem of series and an evaluation of a cubic harmonic series, which is also given in the book, (Almost) Impossible Integrals, Sums, and Series. For a different approach, note that the limit can be brought to the form, $$\displaystyle \int_0^1 \frac{\log (1-x) \log ^2(x)}{(1-x) x} \textrm{d}x$$, where we notice easily that behind the scene there is a classical Euler sums, a well-known one!

A third note: a similar strategy, with some more machinery, has been used in this answer https://math.stackexchange.com/q/3531956.

The work will be turned into an article soon.

Different approach

We have

$$S=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=-\frac12\int_0^1\frac{\ln^2x\ln(1+x)}{x(1+x)}\ dx$$ $$=\frac12\underbrace{\int_0^1\frac{\ln^2x\ln(1+x)}{1+x}\ dx}_{I}-\frac12\underbrace{\int_0^1\frac{\ln^2x\ln(1+x)}{x}\ dx}_{\frac74\zeta(4)}$$

For $$I$$, start with the algebraic identity $$a^2b=\frac13a^3-\frac13b^3+ab^2-\frac13(a-b)^3$$

where if we set $$a=\ln x$$ and $$b=\ln(1+x)$$ we have

$$I=\int_0^1\frac{\ln^2x\ln(1+x)}{1+x}\ dx$$ $$=\frac13\underbrace{\int_0^1\frac{\ln^3x}{1+x}\ dx}_{I_1}-\frac13\underbrace{\int_0^1\frac{\ln^3(1+x)}{1+x}\ dx}_{I_2}+\underbrace{\int_0^1\frac{\ln x\ln^2(1+x)}{1+x}\ dx}_{I_3}-\frac13\underbrace{\int_0^1\frac{\ln^3\left(\frac{x}{1+x}\right)}{1+x}\ dx}_{I_4}$$

$$I_1=\sum_{n=1}^\infty(-1)^{n-1}\int_0^1 x^{n-1}\ln^3x\ dx=6\sum_{n=1}^\infty\frac{(-1)^n}{n^4}=-\frac{21}4\zeta(4)$$

$$I_2=\frac14\ln^42$$

$$I_3\overset{IBP}{=}-\frac13\int_0^1\frac{\ln^3(1+x)}{x}\ dx=2\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac12\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$

Where the last result follows from 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 $$I_4$$ , let $$\frac{x}{1+x}\to x$$

$$I_4=\int_0^{1/2}\frac{\ln^3x}{1-x}\ dx=-6\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{21}4\ln2\zeta(3)+\frac32\ln^22\zeta(2)-\frac1{12}\ln^42$$

which follows from using the generalization

$$\int_0^{1/2}\frac{\ln^n x}{1-x}\ dx=-\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}(-1)^k k!\operatorname{Li}_{k+1}\left(\frac12\right)$$

which can be found on the same link above ( check $$(3)$$).

Combine these results we get

$$I=4\operatorname{Li_4}\left(\frac12\right)-\frac{15}4\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac{1}{6}\ln^42$$

Giving us

$$S=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$$