How to prove


where $H_n^{(q)}=\sum_{k=1}^n\frac{1}{k^q}$ is the generalized harmonic number, $\operatorname{Li}_a(x)=\sum_{k=1}^\infty\frac{x^k}{k^a}$ is the polylogarithmic function and $\zeta$ is the Riemann zeta function.

This problem was proposed by Cornel and no solution has been submitted yet. I managed to convert it to a double integral but it seems tough to crack. Here is what I did:

Using the integral representation of the polygamma function:

$$\int_0^1\frac{x^n\ln^a(x)}{1-x}dx=-\psi^{(a)}(n+1)=(-1)^a a!\left(\zeta(a+1)-H_n^{(a+1)}\right)$$

With $a=2$ we have


multiply both sides by $\frac{H_n}{2n+1}$ then sum up we get


we have

$$\sum_{n=1}^\infty \frac{y^{2n+1}H_n}{2n+1}=-\int_0^y\frac{\ln(1-x^2)}{1-x^2}dx$$

which follows from integrating $\sum_{n=1}^\infty x^{2n}H_n=-\frac{\ln(1-x^2)}{1-x^2}$ from $x=0$ to $x=y$.




For the inner integral, Mathematica gives


and the integral turned out very complicated. So any good idea how to approach the harmonic series or the integral?

Thank you.

  • $\begingroup$ Maybe something is wrong but: wolframalpha.com/input/?i=7%2F4*zeta%282%29*zeta%283%29-279%2F16*zeta%285%29%2B4%2F3*log%282%29%5E3*zeta%282%29-7*log%282%29%5E2*zeta%283%29%2B53%2F4*log%282%29*zeta%284%29-2%2F15*log%282%29%5E5%2Bpolylog%285%2C1%2F2%29 and wolframalpha.com/input/?i=sum+Harmonic%28n%29%2F%282*n%2B1%29*%28zeta%283%29-Harmonic%28n%2C3%29%29%2Cn%3D1%2C100000 $\endgroup$
    – FDP
    Nov 20, 2019 at 13:06
  • $\begingroup$ Sorry @FDP your comment is not clear. $\endgroup$ Nov 20, 2019 at 16:49
  • $\begingroup$ I think the closed-form in the question is wrong but your expression with integrals for the series is right. $\endgroup$
    – FDP
    Nov 20, 2019 at 17:09
  • $\begingroup$ value of closed form and Series $\endgroup$
    – FDP
    Nov 20, 2019 at 17:17
  • $\begingroup$ Ah got you.. actually i didn't check the closed form. I will check it. $\endgroup$ Nov 20, 2019 at 17:18

2 Answers 2


A second solution in large steps by Cornel Ioan Valean

Let's start with the following useful identity which is easily derived by using recurrence relations and simple rearrangements, manipulations with sums, that is

Let $n$ be a non-negative integer number. Then, we have $$\int_0^1 x^{2n}\frac{\log(1+x)}{1+x}\textrm{d}x$$ $$=\frac{1}{2}H_{2n}^2-2\log(2) H_{2n}+\frac{1}{2}H_{2n}^{(2)}-\frac{1}{4}H_n^2-\frac{1}{4}H_n^{(2)}+\log (2)H_n+\frac{1}{2} \log ^2(2)-\sum_{k=1}^{n-1}\frac{H_k}{2 k+1},$$ where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}$ represents the $n$th generalized harmonic number of order $m$.

By multiplying both sides of the identity above by $1/n^3$ and considering the summation from $n=1$ to $\infty$, we get

$$\sum_{n=1}^{\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{H_{k}}{2 k+1}=\sum_{k=1}^{\infty} \sum_{n=k+1}^{\infty}\frac{1}{n^3}\frac{H_{k}}{2 k+1}=\underbrace{\sum_{k=1}^{\infty}\frac{H_{k}}{2 k+1}\left(\zeta(3)-H_k^{(3)}\right)}_{\text{The desired series}}$$ $$=\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_{2n}^2}{n^3}-2\log(2) \sum_{n=1}^{\infty}\frac{H_{2n}}{n^3}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_{2n}^{(2)}}{n^3}-\frac{1}{4}\sum_{n=1}^{\infty}\frac{H_n^2}{n^3}-\frac{1}{4}\sum_{n=1}^{\infty} \frac{H_n^{(2)}}{n^3}$$ $$+\log (2)\sum_{n=1}^{\infty} \frac{H_n}{n^3}+\frac{1}{2}\log ^2(2)\sum_{n=1}^{\infty}\frac{1}{n^3}-\int_0^1 \frac{\log(1+x)}{1+x}\operatorname{Li}_3(x^2)\textrm{d}x,$$

where we see all the series in the right-hand side are easily reducible to known series which may also be found in the book (Almost) Impossible Integrals, Sums, and Series.

On the other hand, with simple integration by parts, we obtain $$\int_0^1 \frac{\log(1+x)}{1+x}\operatorname{Li}_3(x^2)\textrm{d}x$$ $$=\frac{1}{2}\log^2(2)\zeta(3)-2\int_0^1 \frac{\log^2(1+x)\operatorname{Li}_2(x)}{x}\textrm{d}x-2\int_0^1 \frac{\log^2(1+x)\operatorname{Li}_2(-x)}{x}\textrm{d}x,$$ where the last integrals may be found calculated in the paper The calculation of a harmonic series with a weight $5$ structure, involving the product of harmonic numbers, $H_n H_{2n}^{(2)}$.

A note: The sister of the result above (easy to obtain by recurrence relations and very useful),

$$\int_0^1 x^{2n-1} \frac{\log(1+x)}{1+x}\textrm{d}x$$ $$=2\log(2) H_{2n}-\log(2)H_n+\frac{1}{4}H_n^2+\frac{1}{4}H_n^{(2)}-\frac{1}{2}H_{2n}^2-\frac{1}{2} H_{2n}^{(2)}+\frac{H_{2n}}{2n}-\frac{H_n}{2n} $$ $$ -\frac{1}{2}\log^2(2)+\sum_{k=1}^{n-1}\frac{H_k}{2 k+1}. $$

  • 1
    $\begingroup$ I really like this approach (+1). $\endgroup$ Jan 30, 2020 at 20:50
  • $\begingroup$ @AliShather Thanks. $\endgroup$ Jan 30, 2020 at 21:10
  • $\begingroup$ @AliShadhar I'm curious to ask where you found the integral identity stated in the beginning $\endgroup$
    – Math_Day
    Mar 16 at 1:08
  • $\begingroup$ @Math_Day that's Cornel's solution not mine. $\endgroup$ Mar 19 at 21:37

A first solution in large steps by Cornel Ioan Valean

Using The Master Theorem of Series in the article A master theorem of series and an evaluation of a cubic harmonic series or from the book, (Almost) Impossible Integrals, Sums, and Series, pages $288$-$289$, where we set $\mathcal{M}(k)=H_{2k}$, $\displaystyle m(k)=H_{2k}-H_{2k-2}=\frac{1}{2k}+\frac{1}{2k-1}$, we get

$$\sum_{k=1}^{\infty}\frac{H_{2k}}{(k+1)(k+n+1)}$$ $$=\frac{1}{4}\frac{H_n^2}{n}-\log(2)\frac{H_n}{n}+2\log(2)\frac{H_{2n}}{n}+\frac{1}{4}\frac{H_n^{(2)}}{n}-\frac{4\log(2)}{2n+1}+\frac{1}{n}\sum_{k=1}^n \frac{H_k}{2k+1}.$$

Multiplying both sides of the result above by $1/n^2$ and considering the sum from $n=1$ to $\infty$, we have $$\frac{5}{2}\zeta(4)-\frac{1}{2}\zeta(2)\zeta (3)-4\log (2)\zeta(2)+4\sum _{n=1}^{\infty } \frac{H_n}{n(2n-1)}-2\sum _{n=1}^{\infty } \frac{H_n}{n^2}-\sum _{n=1}^{\infty } \frac{H_n}{n^3}$$ $$+\frac{1}{2}\sum _{n=1}^{\infty } \frac{H_n}{n^4}+\zeta(2)\sum _{n=1}^{\infty } \frac{H_{2 n}}{n^2}-\underbrace{\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^3}}_{\text{A tough series}}$$ $$=16 \log (2)-16 \log ^2(2)-4\log (2)\zeta(2)+\frac{1}{4}\sum _{n=1}^{\infty } \frac{H_n^2}{n^3}+\sum _{n=1}^{\infty } \frac{H_n}{(2 n+1) n^3}-\log (2)\sum _{n=1}^{\infty } \frac{ H_n}{n^3}$$ $$+2 \log (2)\sum _{n=1}^{\infty } \frac{ H_{2 n}}{n^3}+\frac{1}{4}\sum _{n=1}^{\infty } \frac{H_n^{(2)}}{n^3}+\underbrace{\sum _{n=1}^{\infty } \frac{H_n}{2 n+1}\left(\zeta (3)-H_n^{(3)}\right)}_{\text{The desired series}},$$ where in the calculations we used the folowing result from the paper On the calculation of two essential harmonic series with a weight $5$ structure, involving harmonic numbers of the type $H_{2n}$, $$\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n)^3}$$ $$=\frac{307}{128}\zeta(5)-\frac{1}{16}\zeta (2) \zeta (3)+\frac{1}{3}\log ^3(2)\zeta (2) -\frac{7}{8} \log ^2(2)\zeta (3)-\frac{1}{15} \log ^5(2)$$ $$-2 \log (2) \operatorname{Li}_4\left(\frac{1}{2}\right) -2 \operatorname{Li}_5\left(\frac{1}{2}\right),$$ and at the same time we used that by reversing the summation order our desired series is (almost) revealed $$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^3}\sum_{k=1}^n \frac{H_k}{2k+1}=\sum _{n=1}^{\infty } \frac{H_n}{2 n+1}\left(\zeta (3)-H_n^{(3)}+\frac{1}{n^3}\right).$$

The rest of the series are well-known (or immediately reducible to well-known series), and all of them may also be found calculated in the book, (Almost) Impossible Integrals, Sums, and Series.

A note: By using the same procedure, we may calculate the series $$\sum _{n=1}^{\infty } \frac{H_n}{2 n+1}\left(\zeta (2)-H_n^{(2)}\right),$$ or other versions if we know the resulting series after applying this strategy.

  • 1
    $\begingroup$ (+1) for the nice work. $\endgroup$ Jan 30, 2020 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.