# Proving that $~\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{k^2}~H_k~H_{k-1}=\frac{3}{16}~\zeta(4)$

To show that $$\sum\limits_{k=2}^{\infty} \frac{(-1)^{k}}{k^{2}} \, \left(1+\frac{1}{2}+...+\frac{1}{k}\right) \cdot \left(1+\frac{1}{2}+...+\frac{1}{k-1}\right) = \frac{3}{16}\zeta(4).$$

I came across this when trying to solve a problem from the current edition of the American Mathematical Monthly. Is there some easy way to show this? I checked numerically that this series does converge to the value of $\frac{3}{16}\zeta(4)$.

Note: An alternate form, with $H_{n}$ being the harmonic numbers, is: $$\sum\limits_{k=2}^{\infty} \frac{(-1)^{k}}{k^{2}} \, H_{k} \, H_{k-1} = \frac{3}{16}\zeta(4).$$

• algo.inria.fr/flajolet/Publications/FlSa98.pdf has several techniques that might be useful, but I haven't finished reading it. Either way, it's a fun read :D Commented Jul 28, 2017 at 20:30
• My go-to reference for this kind of stuff are the Borwein brothers. Check out this paper: ams.org/journals/proc/1995-123-04/S0002-9939-1995-1231029-X/… specifically, the theorem and its corollaries under "Main Results" on page 2. My money is on manipulating the integral to give you that $H_kH_{k-1}$ combination. In particular check out the integrals that follow under equation (16) on page 7, which give the non-alternating sum version of your identity. Commented Jul 28, 2017 at 20:48
• Link for the AMM problem. Here's an old spoiler blog post. But trust me, the problem in that form is much simpler than what I did in that blog. :P
– r9m
Commented Jul 28, 2017 at 21:56
• @r9m impressive work! Commented Jul 28, 2017 at 23:33
• @Alex Meiburg,@ Alex R: Thanks a lot for all the references, they are all very interesting. The link zerocollar.blogspot.in/2015/05/… shared by r9m pretty much solves it. Commented Jul 29, 2017 at 2:51

This is an opportunity to make a tribute to Pieter J. de Doelder (1919-1994) from Eindhoven University of Technology, who evaluated in closed form the given series in a somewhat famous paper (p. 132-133 2.3) (1991). One may start by using the following identity coming from the Cauchy product, $$\ln^2(1+x) =2\sum_{n=1}^\infty (-1)^{n-1} \frac{H_n}{n+1} \:x^{n+1}$$ giving $$\int_0^1\frac{\ln(1-x)\ln^2(1+x)}{x} \:dx=2\sum_{n=1}^\infty (-1)^{n-1} \frac{H_n}{n+1} \:\int_0^1 x^{n}\ln(1-x)\:dx,$$ then using the standard evaluation $$\int_0^1 x^{n}\ln(1-x)\:dx =-\frac{H_{n+1}}{n+1},\quad n\ge0,$$ one gets

$$\int_0^1\frac{\ln(1-x)\ln^2(1+x)}{x}\:dx=2\sum\limits_{n=2}^{\infty} (-1)^{n-1} \frac{H_n H_{n-1}}{n^{2}}. \tag1$$

Here are the main steps which de Doelder took to evaluate the related integral.

We clearly have \begin{align} \int_0^1\ln^3\left(\frac{1+x}{1-x}\right)\:\frac{dx}{x}&=\int_0^1\frac{\ln^3\left(1+x\right)}{x}\:dx-3\int_0^1\frac{\ln(1-x)\ln^2(1+x)}{x}\:dx \\\\&+3\int_0^1\frac{\ln^2(1-x)\ln(1+x)}{x}\:dx-\int_0^1\frac{\ln^3\left(1-x\right)}{x}\:dx \end{align} and \begin{align} \int_0^1\frac{\ln^3\left(1-x^2\right)}{x}\:dx&=\int_0^1\frac{\ln^3\left(1+x\right)}{x}\:dx+3\int_0^1\frac{\ln(1-x)\ln^2(1+x)}{x}\:dx \\\\&+3\int_0^1\frac{\ln^2(1-x)\ln(1+x)}{x}\:dx+\int_0^1\frac{\ln^3\left(1-x\right)}{x}\:dx, \end{align} substracting the two equalities, \begin{align} 6\!\!\int_0^1\!\frac{\ln(1-x)\ln^2(1+x)}{x}dx&=\!\int_0^1\!\frac{\ln^3\left(1-x^2\right)}{x}dx-\!\int_0^1\!\!\ln^3\left(\frac{1+x}{1-x}\right)\frac{dx}{x}-2\!\int_0^1\!\frac{\ln^3\left(1-x\right)}{x}dx \\\\&=I_1-I_2-2I_3. \end{align} It is easy to obtain \begin{align} I_1=\int_0^1\!\frac{\ln^3\left(1-x^2\right)}{x}dx&=\frac12 \int_0^1\!\frac{\ln^3\left(1-u\right)}{u}du \quad (u=x^2) \\&=\frac12 \int_0^1\!\frac{\ln^3 v}{1-v}dv \quad (v=1-u) \\&=\frac12 \sum_{n=0}^\infty \int_0^1\!v^n\ln^3 v\:dv \\&=-3\sum_{n=1}^\infty \frac1{n^4} \\&=-\frac{\pi^4}{30}, \end{align} similarly \begin{align} I_3=\int_0^1\!\frac{\ln^3\left(1-x\right)}{x}dx=-\frac{\pi^4}{15}. \end{align} By the change of variable, $u=\dfrac{1-x}{1+x}$, one has $\dfrac{dx}{x}=\dfrac{-2\:du}{1-u^2}$ getting \begin{align} I_2=\int_0^1\!\!\ln^3\left(\frac{1+x}{1-x}\right)\frac{dx}{x}&=-2\int_0^1\!\frac{\ln^3 u}{1-u^2}du \\&=-2\sum_{n=0}^\infty \int_0^1\!u^{2n}\ln^3u\:dv \\&=12\sum_{n=0}^\infty \frac1{(2n+1)^4} \\&=\frac{\pi^4}{8}. \end{align} Then,

$$\int_0^1\frac{\ln(1-x)\ln^2(1+x)}{x}\:dx=-\frac{\pi^4}{240} \tag2$$

and

$$\sum\limits_{n=2}^{\infty} (-1)^{n} \frac{H_n H_{n-1}}{n^{2}}= \frac{3}{16}\zeta(4)=\frac{\pi^4}{480},\tag3$$

as announced.

• thanks! This integral is exactly the problem proposed in AMM this month. Would it be wrong to publicly keep this solution up here in that case? Commented Jul 29, 2017 at 12:39
• (+1) Nice solution .. !! :) Elementary and elegant as usual!
– r9m
Commented Jul 29, 2017 at 18:25
• @Aritro Pathak You are welcome! It's no big deal, this monthly problem is not original, it is known for years. Commented Jul 29, 2017 at 18:25
• Ingenious! Regarding the question of motivation for the solution, I suppose you do have to memorize identities like the first and third, is that right?
– Hans
Commented Jul 29, 2017 at 23:06
• @Hans Right, and if you don't memorize it, you know it does exist somewhere. Thanks. Commented Jul 30, 2017 at 0:31

Although I have seen too few proofs in this field to be able to compare, this approach might be interesting.

We transform the sum to a fourfold integral which Mathematica can solve immediately. I hope it should be possible to solve the integral "mathematically" as well, which would then complete the proof.

We have to calculate

$$s=\sum _{k=2}^{\infty } \frac{(-1)^k}{k^2} H(k) H(k-1)$$

Writing

$$\frac{1}{n^2}=\int_0^1 \frac{1}{x}\,dx \int_0^x y^{n-1} \, dy$$

$$\frac{1}{n}=\int_0^1 r^{n-1} \, dr$$

and

$$H(k)=\sum _{n=1}^k \frac{1}{n}=\int_0^1 \left(\sum _{n=1}^k r^{n-1}\right) \, dr=\int_0^1 \frac{1-r^k}{1-r} \, dr$$

the sum $s$ below the integrals becomes

$$si=\frac{1}{x(1-r)(1-s)}\sum _{k=2}^{\infty } (-1)^k \left(1-r^k\right) \left(1-s^{k-1}\right) y^{k-1}$$

Which evaluates to

$$si = \frac{y \left(r^2 s^2 y+r^2 s-r^2 y-r^2-r s^2 y+r y-s+1\right)}{(1-r) (1-s) x (y+1) (r y+1) (s y+1) (r s y+1)}$$

Now the integral to be evaluated is

$$s4 = \int _0^1 dx\int _0^x dy\int _0^1 dr\int _0^1 ds \; si$$

Mathematica finds immediately

$$s4 = \frac{\pi ^4}{480}$$

Since

$$\zeta (4)=\frac{\pi ^4}{90}$$

and

$$\frac{90}{480} = \frac{3}{16}$$

we have finally

$$s = \frac{3}{16} \zeta(4)$$

• Thank you for this point of view! (+1) Is it possible to have a screenshot of your instructions in Mathematica? Thanks. Commented Jul 30, 2017 at 7:44
• @Olivier Oloa No screenshot necessary. Using the integrand $si$ as defined above it's just this statement Integrate[si, {x, 0, 1}, {y, 0, x}, {r, 0, 1}, {s, 0, 1}]. To my big surprise the result pi^4/480 was returned within a second. So if I had not known the result in advance I would have found it heuristically. This is not replacing a strict proof, of course. Commented Jul 30, 2017 at 8:49
• Maple can't integrate, so it's not that trivial ;-( Commented Jan 8, 2019 at 0:35
• @Diger I have never used maple but did you try different versions of maple? I keep several older versions of Mathematica because some are better in integrating than others. Commented Jan 9, 2019 at 8:27
• Unfortunately not; just tried 2018. Commented Jan 9, 2019 at 14:27

Bonus

$$S=\sum_{k=2}^\infty\frac{(-1)^k}{k^2}H_kH_{k-1}=\sum_{k=1}^\infty\frac{(-1)^k}{k^2}H_kH_{k-1}=\sum_{k=1}^\infty\frac{(-1)^kH_k^2}{k^2}-\sum_{k=1}^\infty\frac{(-1)^kH_k}{k^3}$$

Since $$S=\frac3{16}\zeta(4)$$ and $$\sum_{k=1}^\infty\frac{(-1)^kH_k}{k^3}$$ $$=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$$

then

$$\sum_{n=1}^{\infty}\frac{(-1)^kH_k^2}{k^2}=2\operatorname{Li}_4\left(\frac12\right)-\frac{41}{16}\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$