I found the following formula

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

and it is cited that Euler proved the formula above , but how ?

Do there exist other proofs ?

Can we have a general formula for the alternating form

$$\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^q}$$

  • $\begingroup$ (at)Zaid Alyafeai The community here might be interested in my solution attempt for the alternating form. $\endgroup$ Sep 29, 2017 at 22:32
  • $\begingroup$ The post is rather old but interesting of course! Do you have the source where Euler is mentioned as the person proving it? And where can one see this result - I mean where did you find this? Thank you :-) $\endgroup$
    – Math-fun
    Dec 7, 2018 at 16:57

7 Answers 7


$$ \begin{align} &\sum_{j=0}^k\zeta(k+2-j)\zeta(j+2)\\ &=\sum_{m=1}^\infty\sum_{n=1}^\infty\sum_{j=0}^k\frac1{m^{k+2-j}n^{j+2}}\tag{1}\\ &=(k+1)\zeta(k+4) +\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{m^2n^2} \frac{\frac1{m^{k+1}}-\frac1{n^{k+1}}}{\frac1m-\frac1n}\tag{2}\\ &=(k+1)\zeta(k+4) +\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{nm^{k+2}(n-m)}-\frac1{mn^{k+2}(n-m)}\tag{3}\\ &=(k+1)\zeta(k+4) +2\sum_{m=1}^\infty\sum_{n=m+1}^\infty\frac1{nm^{k+2}(n-m)}-\frac1{mn^{k+2}(n-m)}\tag{4}\\ &=(k+1)\zeta(k+4) +2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{(n+m)m^{k+2}n}-\frac1{m(n+m)^{k+2}n}\tag{5}\\ &=(k+1)\zeta(k+4)\\ &+2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^{k+3}n}-\frac1{(m+n)m^{k+3}}\\ &-2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m(n+m)^{k+3}}+\frac1{n(n+m)^{k+3}}\tag{6}\\ &=(k+1)\zeta(k+4) +2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}} -4\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1{n(n+m)^{k+3}}\tag{7}\\ &=(k+1)\zeta(k+4) +2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}} -4\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac1{nm^{k+3}}\tag{8}\\ &=(k+1)\zeta(k+4) +2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}} -4\sum_{n=1}^\infty\sum_{m=n}^\infty\frac1{nm^{k+3}}+4\zeta(k+4)\tag{9}\\ &=(k+5)\zeta(k+4) +2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}} -4\sum_{m=1}^\infty\sum_{n=1}^m\frac1{nm^{k+3}}\tag{10}\\ &=(k+5)\zeta(k+4) +2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}} -4\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}\tag{11}\\ &=(k+5)\zeta(k+4) -2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}\tag{12} \end{align} $$ Letting $q=k+3$ and reindexing $j\mapsto j-1$ yields $$ \sum_{j=1}^{q-2}\zeta(q-j)\zeta(j+1) =(q+2)\zeta(q+1)-2\sum_{m=1}^\infty\frac{H_m}{m^q}\tag{13} $$ and finally $$ \sum_{m=1}^\infty\frac{H_m}{m^q} =\frac{q+2}{2}\zeta(q+1)-\frac12\sum_{j=1}^{q-2}\zeta(q-j)\zeta(j+1)\tag{14} $$


$\hphantom{0}(1)$ expand $\zeta$
$\hphantom{0}(2)$ pull out the terms for $m=n$ and use the formula for finite geometric sums on the rest
$\hphantom{0}(3)$ simplify terms
$\hphantom{0}(4)$ utilize the symmetry of $\frac1{nm^{k+2}(n-m)}+\frac1{mn^{k+2}(m-n)}$
$\hphantom{0}(5)$ $n\mapsto n+m$ and change the order of summation
$\hphantom{0}(6)$ $\frac1{mn}=\frac1{m(m+n)}+\frac1{n(m+n)}$
$\hphantom{0}(7)$ $H_m=\sum_{n=1}^\infty\frac1n-\frac1{n+m}$ and use the symmetry of $\frac1{m(n+m)^{k+3}}+\frac1{n(n+m)^{k+3}}$
$\hphantom{0}(8)$ $m\mapsto m-n$
$\hphantom{0}(9)$ subtract and add the terms for $m=n$
$(10)$ combine $\zeta(k+4)$ and change the order of summation
$(11)$ $H_m=\sum_{n=1}^m\frac1n$
$(12)$ combine sums

  • 4
    $\begingroup$ This works for any integer $q\ge2$. $\endgroup$
    – robjohn
    Aug 18, 2013 at 16:03
  • $\begingroup$ Your answer is Euler's original proof? Can we start with $\sum_{j=1}^{k-2}\zeta(k-j)\zeta(j+1)$ at the first line? $\endgroup$
    – user91500
    Dec 15, 2013 at 11:11
  • 2
    $\begingroup$ @ALGEAN: I don't know how Euler did it, so I can't answer your first question. I don't see why you couldn't start with that as the first line. $\endgroup$
    – robjohn
    Dec 15, 2013 at 11:31
  • $\begingroup$ @robjohn The community here might be interested in my solution attempt for the alternating form. $\endgroup$ Sep 29, 2017 at 22:30

Answering the first part of the question for $q$ odd we recall from the following MSE post the identity: $$ H_n = - \frac{1}{2\pi i} \int_{-1/2-i\infty}^{-1/2+i\infty} \zeta(1-s) \frac{\pi}{\sin(\pi s)}\frac{1}{n^s} ds.$$ The proof at the above cited post is sound and I will merely refer to it here since otherwise we would just include it verbatim.

This gives the formula for your sum: $$\sum_{n\ge 1} \frac{H_n}{n^q} = - \frac{1}{2\pi i} \int_{-1/2-i\infty}^{-1/2+i\infty} \zeta(1-s) \frac{\pi}{\sin(\pi s)} \zeta(q+s) ds.$$

Now shift this integral to the left to the line $\Re(s) = -1/2-(q-1),$ getting $$\sum_{n\ge 1} \frac{H_n}{n^q} = \rho_1 - \sum_{k=1}^{q-2} \zeta(1+k) (-1)^k \zeta(q-k) - \frac{1}{2\pi i} \int_{-1/2-(q-1)-i\infty}^{-1/2-(q-1)+i\infty} \zeta(1-s) \frac{\pi}{\sin(\pi s)} \zeta(q+s) ds$$ where $$\rho_1 = \operatorname{Res}\left( -\zeta(1-s) \frac{\pi}{\sin(\pi s)} \zeta(q+s); s=-(q-1)\right).$$

Make the substitution $t=s+(q-1)$ in the integral to get (not including the minus sign in front) $$ \frac{1}{2\pi i}\int_{-1/2-i\infty}^{-1/2+i\infty} \zeta(1-(t-(q-1))) \frac{\pi}{\sin(\pi (t-(q-1))} \zeta(q+t-(q-1)) dt.$$ For $q$ odd this simplifies to $$ \frac{1}{2\pi i}\int_{-1/2-i\infty}^{-1/2+i\infty} \zeta(q-t) \frac{\pi}{\sin(\pi t)} \zeta(t+1) dt.$$ Now make another substitution, namely $v=-t$, to get $$ \frac{1}{2\pi i}\int_{1/2+i\infty}^{1/2-i\infty} \zeta(q+v) \frac{\pi}{\sin(\pi v)} \zeta(1-v) dv =-\frac{1}{2\pi i}\int_{1/2-i\infty}^{1/2+i\infty} \zeta(q+v) \frac{\pi}{\sin(\pi v)} \zeta(1-v) dv$$ where the minus on the sine term cancels the one on the differential. Finally shift this integral to the line $\Re(v) = -1/2$ to obtain $$\rho_2 - \frac{1}{2\pi i}\int_{-1/2-i\infty}^{-1/2+i\infty} \zeta(q+v) \frac{\pi}{\sin(\pi v)} \zeta(1-v) dv = \rho_2 + \sum_{n\ge 1} \frac{H_n}{n^q}$$ where $$\rho_2 = \operatorname{Res}\left(- \zeta(1-v) \frac{\pi}{\sin(\pi v)} \zeta(q+v); v=0\right).$$ We have shown that $$\sum_{n\ge 1} \frac{H_n}{n^q} = \rho_1 - \sum_{k=1}^{q-2} \zeta(1+k) (-1)^k \zeta(q-k) - \left(\rho_2 + \sum_{n\ge 1} \frac{H_n}{n^q}\right).$$ This gives $$ \sum_{n\ge 1} \frac{H_n}{n^q} = \frac{1}{2} (\rho_1-\rho_2) - \frac{1}{2} \sum_{k=1}^{q-2} \zeta(1+k) (-1)^k \zeta(q-k).$$ To conclude introduce $$ W(s) = -\zeta(1-s) \frac{\pi}{\sin(\pi s)} \zeta(q+s).$$ This implies that $$ W(-s-(q-1)) = -\zeta(s+q) \frac{\pi}{\sin(\pi (-s-(q-1)))} \zeta(1-s) = - W(s)$$ because $q$ is odd. Now $$\rho_2 = \frac{1}{2\pi i} \int_{|s|=1/2} W(s) ds.$$ Put $s = -t -(q-1)$ and note that this does not change the counterclockwise orientation of the circle induced by the first integral to get $$ -\frac{1}{2\pi i} \int_{|-t-(q-1)|=1/2} W(-t-(q-1)) dt = \frac{1}{2\pi i} \int_{|-t-(q-1)|=1/2} W(t) dt = \rho_1$$ because $|-t-(q-1)|=|(-1)(t+(q-1))|=|t-(-(q-1))|.$ The conclusion is that $$ \sum_{n\ge 1} \frac{H_n}{n^q} = -\frac{1}{2} \sum_{k=1}^{q-2} \zeta(1+k) (-1)^k \zeta(q-k)$$ for $q$ odd.

Addendum. Sun Apr 27 23:57:35 CEST 2014 I don't quite see why I didn't simply evaluate the residues $\rho_1$ and $\rho_2$ as these are both easy. This does not affect the correctness of the argument.

Addendum. Sun Nov 9 23:33:24 CET 2014 In fact the equality of the two residues follows by inspection. In retrospect it appears I wanted to avoid working with the two double poles and keep everything within the limits of pen and paper.

  • 1
    $\begingroup$ Wonderful answer .How have you thought of just a proof ? $\endgroup$ Aug 17, 2013 at 0:42
  • $\begingroup$ By a curious coincidence this question indirectly referred to another one I did yesterday, which is what motivated me to give it a try. $\endgroup$ Aug 17, 2013 at 0:48
  • $\begingroup$ Interestingly we discuss similar methods here integralsandseries.prophpbb.com/topic136.html , if you are interested you could join ! $\endgroup$ Aug 17, 2013 at 0:51
  • $\begingroup$ (+1) nice work. Here is a related technique. $\endgroup$ Aug 17, 2013 at 1:19
  • $\begingroup$ Thanks for the pointer and the kind remark, I had seen your work before. I do have quite a few of these (i.e. related subject matter) which you can find in my profile. $\endgroup$ Aug 17, 2013 at 1:28

When $q$ is odd and greater than $1$, one can show $$ \sum_{n=1}^{\infty} \frac{H_{n}}{n^{q}} = \frac{1}{2} \sum_{k=1}^{q-2} (-1)^{k-1} \zeta(k+1) \zeta(q-k)$$

by replacing $H_{n}$ with the integral representation

$$ H_{n} = \int_{0}^{1} \frac{1-x^{n}}{1-x} \, dx \ ,$$

switching the order of integration and summation, and then repeatedly integrating by parts.

This result is also derived in Marko Riedel's answer using a different approach.

$$ \begin{align} \sum_{n=1}^{\infty} \frac{H_{n}}{n^{q}} &= \sum_{n=1}^{\infty} \frac{1}{n^{q}} \int_{0}^{1} \frac{1-x^{n}}{1-x} \, dx \\ &= \int_{0}^{1} \frac{1}{1-x} \sum_{n=1}^{\infty} \frac{1-x^{n}}{n^{q}} \, dx \\ &= \int_{0}^{1} \frac{\zeta(q)- \text{Li}_{q}(x)}{1-x} \, dx \\ &= - \Big(\zeta(q) - \text{Li}_{q}(x) \Big) \ln(1-x) \Bigg|^{1}_{0} - \int_{0}^{1} \frac{\log(1-x) \text{Li}_{q-1}(x)}{x} \, dx \\ &= -\color{#C00000} {\int_{0}^{1} \frac{\log(1-x) \text{Li}_{q-1}(x)}{x} \, dx} \\ &= \text{Li}_{2}(x) \text{Li}_{q-1}(x) \Bigg|^{1}_{0} - \int_{0}^{1} \frac{\text{Li}_{2}(x) \text{Li}_{q-2}(x)}{x} \, dx \\ &= \zeta(2) \zeta(q-1) - \int_{0}^{1} \frac{\text{Li}_{2}(x) \text{Li}_{q-2}(x)}{x} \, dx \\ &= \zeta(2) \zeta(q-1) - \text{Li}_{3}(x) \text{Li}_{q-2}(x) \Bigg|^{1}_{0} + \int_{0}^{1} \frac{\text{Li}_{3}(x)\text{Li}_{q-3}(x) }{x} \, dx \\ &= \zeta(2) \zeta(q-1) - \zeta(3) \zeta(q-2) + \int_{0}^{1} \frac{\text{Li}_{3}(x)\text{Li}_{q-3}(x) }{x} \, dx \\&= \zeta(2) \zeta(q-1) - \zeta(3) \zeta(q-2) + \zeta(4) \zeta(q-3) - \int_{0}^{1} \frac{\text{Li}_{4}(x) \text{Li}_{4-q}(x)}{x} \, dx \\ &=\zeta(2) \zeta(q-1) - \zeta(3) \zeta(q-2) + \zeta(4) \zeta(q-3) - \ldots + \zeta(q-1) \zeta(2) - \int_{0}^{1} \frac{\text{Li}_{q-1}(x) \text{Li}_{1}(x)}{x} \, dx \\ &= \sum_{k=1}^{q-2} (-1)^{k-1} \zeta(k+1) \zeta(q-k) + \color{#C00000}{\int_{0}^{1} \frac{\log(1-x) \text{Li}_{q-1}(x)}{x} \, dx} \end{align}$$

Therefore, if $q$ is odd,

$$\sum_{n=1}^{\infty} \frac{H_{n}}{n^{q}} = - \int_{0}^{1} \frac{\log(1-x) \text{Li}_{q-1}(x)}{x} \, dx = \frac{1}{2} \sum_{k=1}^{q-2} (-1)^{k-1} \zeta(k+1) \zeta(q-k).$$


Note that,

$\displaystyle \int_{0}^{1} x^{n-1} \mathrm{d}x = \dfrac{1}{n}$

Differentiating w.r.t. to $n$, $(p-1)$ times, we get,

$\displaystyle \dfrac{1}{n^{p}} = \dfrac{(-1)^{p-1}}{(p-1)!} \int_{0}^{1} x^{n-1} [\ln(x)]^{p-1} \mathrm{d}x$

$\displaystyle \implies \text{S} = \sum_{n=1}^{\infty} \dfrac{H_{n}}{n^{p}} = \dfrac{(-1)^{p-1}}{(p-1)!} \int_{0}^{1} [\ln(x)]^{p-1} \sum_{n=1}^{\infty} H_{n} x^{n-1} \mathrm{d}x $

Since $\displaystyle \sum_{n=1}^{\infty} H_{n} x^{n} = -\dfrac{\ln(1-x)}{1-x} $, we get,

$\displaystyle \text{S} = \dfrac{(-1)^{p}}{(p-1)!} \int_{0}^{1}\dfrac{[\ln(x)]^{p-1} \cdot \ln(1-x) }{x(1-x)} \mathrm{d}x $

Recall the Beta Function $\displaystyle \operatorname{B}(a,b) = \int_{0}^{1} x^{a-1} (1-x)^{b-1} \mathrm{d}x = \dfrac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)}$

$\displaystyle \implies \text{S} = \dfrac{(-1)^{p}}{(p-1)!} \lim_{a \to 0^+} \lim_{b \to 0^+} \left(\dfrac{{\partial}^{p-1}}{\partial a^{p-1}} \left( \dfrac{\partial}{\partial b} \operatorname{B}(a,b) \right)\right) $

$\therefore \displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n}}{n^{p}} = \left(1+\dfrac{p}{2} \right)\zeta(p+1)-\dfrac{1}{2}\sum_{k=1}^{p-2}\zeta(k+1)\zeta(p-k)$.

This is valid for any integer $p \geq 2$.

  • $\begingroup$ Ingenious derivation. $\endgroup$
    – Mr Pie
    Aug 29, 2021 at 15:47

Although this problem is from April 2013 I would like to take it up and try to complete the answer turning to the question

"Can we have a general formula for the alternating form?"

$$S_a(q) = \sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^q}$$

By inspecting the first various expressions I have made the following guess for the alternating series for even $q = 2, 4, ...$

$$S_a(q=2,4,...) = c(q)\frac{ \zeta (q+1)}{2^{q+1}}-\sum _{k=1}^{\frac{q}{2}-1} \left(1-\frac{1}{2^{q-2 k-1}}\right) \zeta (2 k+1) \zeta (q-2 k)\tag{1}$$

Here $c(q)$ are coefficients. The first 10 entries are

$$c(2,4,..,20) = \{5,59,377,2039,10229,49139,229361,1048559,4718573,20971499\}\tag{1a}$$

This sequence is not contained in https://oeis.org and I could not find a formula up to now.

For odd $q$ Mathematica returns a seemingly simple pattern

$$S_a(q=1)= \frac{\pi ^2}{12}-\frac{\log ^2(2)}{2}\tag{2a}$$

$$S_a(q=3,5,...)= \gamma \left(1-\frac{1}{2^{q-1}}\right) \zeta (q)-\;{_aF}_b^{reg}(q)\tag{2b}$$

where $\gamma$ is the Euler gamma, and ${_ aF}_b^{reg}(q)$ is the partial derivative of the regularized hypergeometric function with the parameter sets $a$ and $b$ with repect to the last parameter in $b$ taken at the argument -1.

I still need to understand this function better before posting it here. Most probably it hides a pattern similar to that of (1).


After having completed the entry up to this point I found that the case of odd $q$ has already been treated extensively in Calculating alternating Euler sums of odd powers in March 2017.

Using these results we can easily identify the coefficients (1a) as

$$c(q) = q \left(2^q-1\right)-1$$


We have: \begin{eqnarray} \sum\limits_{n=1}^\infty \frac{H_n}{n^q} &=& \sum\limits_{n=1}^\infty \frac{H_n}{(n+1)^q} + \zeta(q+1) \\ &=& 1/2 \left(q \zeta(q+1) - \sum\limits_{j=1}^{q-2} \zeta(j+1) \zeta(q-j) \right)+ \zeta(q+1) \end{eqnarray} where in the last line we used the result given in the answer to question Closed form expressions for harmonic sums .


Partial solution:

I am going to prove

$$\sum_{k=1}^\infty\frac{H_k}{k^n}=\frac12\sum_{i=1}^{n-2}(-1)^{i-1}\zeta(n-i)\zeta(i+1),\quad n=3,5,7, ...$$

We have

$$\int_0^1x^{k-1}\operatorname{Li}_n(x)\ dx\overset{IBP}{=}(-1)^{n-1}\frac{H_k}{k^n}-\sum_{i=1}^{n-1}(-1)^i\frac{\zeta(n-i+1)}{k^i}$$

Divide both sides by $k$ then consider the summation from $k=1$ to $\infty$ we have

$$\int_0^1\frac{\operatorname{Li}_n(x)}{x}\sum_{k=1}^\infty\frac{x^k}{k}\ dx=(-1)^{n-1}\sum_{k=1}^\infty\frac{H_k}{k^{n+1}}-\sum_{i=1}^{n-1}(-1)^i\zeta(n-i+1)\sum_{k=1}^\infty\frac1{k^{i+1}}$$

$$\small{-\int_0^1\frac{\operatorname{Li}_n(x)\ln(1-x)}{x}\ dx=(-1)^{n-1}\sum_{k=1}^\infty\frac{H_k}{k^{n+1}}-\sum_{i=1}^{n-1}(-1)^i\zeta(n-i+1)\zeta(i+1)}\tag1$$


$$-\int_0^1\frac{\operatorname{Li}_n(x)\ln(1-x)}{x}\ dx=-\sum_{k=1}^\infty\frac1{k^n}\int_0^1 x^{k-1}\ln(1-x)\ dx=\sum_{k=1}^\infty\frac{H_k}{k^{n+1}}\tag2$$

Plug (2) in (1) we get


Let $n-1\mapsto n$ to get


So clearly for odd $n\geq3$ we have


set $n=2m+1$

$$\sum_{k=1}^\infty\frac{H_k}{k^{2m+1}}=-\frac12\sum_{i=1}^{2m-1}(-1)^i\zeta(2m+1-i)\zeta(i+1),\quad m=1,2,3,...$$


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.