How to prove that $$\sum_{n=1}^{\infty}\frac{(H_n)^2}{n^3}=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)$$ $H_n$ denotes the harmonic numbers.

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    $\begingroup$ Typo: you meant $H_n = \sum_{k=1}^n \frac 1k$. $\endgroup$ – Stephen Montgomery-Smith Nov 7 '13 at 5:40
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    $\begingroup$ You can use the integral representation $B(p,q)$. Also, see here for related problem and techniques. $\endgroup$ – Mhenni Benghorbal Nov 7 '13 at 5:49
  • $\begingroup$ Generally speaking, $H_n\simeq\ln n+\gamma$, and $$\begin{align}\sum_{n=1}^\infty\frac{H_n^a}{n^b}&\simeq\sum_{n=1}^\infty\frac{\ln^an}{n^b}=\left|\zeta^{(a)}(b)\right|\simeq\\\\\\\\\\&\simeq\int_1^\infty\frac{\ln^ax}{x^b}dx=\frac{a!}{(b-1)^{a+1}}\end{align}$$ where the simpler formula $n!=\int_1^\infty\frac{\ln^nx}{x^2}dx$ can be obtained by a simple change in variable $t=\frac1x$ applied to Euler's first integral expression for the $\Gamma$ function: $n!=\int_0^1(-\ln x)^ndx=\int_0^1\ln^n\frac1xdx$. $\endgroup$ – Lucian Nov 9 '13 at 0:47
  • $\begingroup$ You should add explicitly what is $\large H_{n}$. $\endgroup$ – Felix Marin Nov 9 '13 at 6:19
  • $\begingroup$ Also, try and simply write the first few terms of each $\zeta$ from each product and simply multiply them, then group and order them, and notice the general form of the terms of each such product. $\endgroup$ – Lucian Nov 9 '13 at 7:04

One systematic way of proving identities like this is to write everything in terms of multiple harmonic sums and multiple zeta values . For integers $s_1,\ldots,s_k,n\geq 1$, we define the multiple harmonic sum (or MHS) $$ H_n(s_1,\ldots,s_k):=\sum_{n\geq n_1>\ldots>n_k\geq 1}\frac{1}{n_1^{s_1}\ldots n_k^{s_k}}\in\mathbb{Q}. $$ The MHS $H_n(1)$ is what you call $H_n$. We also define the multiple zeta value (or MZV) $$ \zeta(s_1,\ldots,s_k):=\sum_{n_1>\ldots>n_k\geq 1}\frac{1}{n_1^{s_1}\ldots n_k^{s_k}}\in\mathbb{R}, $$ where we need $s_1\geq 2$ to ensure convergence.

The following relationship between MHS and MZV is easy to check, and will be useful: $$ \sum_{n=1}^\infty \frac{H_n(s_1,\ldots,s_k)}{n^s}=\zeta(s,s_1,\ldots,s_k)+\zeta(s+s_1,s_2,\ldots,s_k). $$

MHS's and MZV's satisfy what's called a quasi-shuffle identity. One instance of this is given by the following: \begin{eqnarray*} H_n(1)^2&=&\left(\sum_{n_1=1}^n\frac{1}{n_1}\right)\left(\sum_{n_2=1}^n\frac{1}{n_2}\right)\\ &=&\left(\sum_{n\geq n_1>n_2}+\sum_{n\geq n_2>n_1}+\sum_{n\geq n_1=n_2}\right)\frac{1}{n_1n_2}\\ &=&2H_n(1,1)+H_n(2). \end{eqnarray*}

Using the relationship between MHS and MZV and the quasi-shuffle identity, the left hand side of your identity equals $$ 2\zeta(3,1,1)+2\zeta(4,1)+\zeta(3,2)+\zeta(5). $$

The expression $\zeta(2)\zeta(3)$ can also be expanded with a quasi-shuffle identity: $$ \zeta(2)\zeta(3)=\zeta(2,3)+\zeta(3,2)+\zeta(5). $$ This means your identity is equivalent to the MZV identity $$ 2\zeta(3,1,1)+2\zeta(4,1)+2\zeta(3,2)+\zeta(2,3)=\frac{3}{2}\zeta(5). $$ This identity is linear and homogeneous (that is, each MZV $\zeta(s_1,\ldots,s_k)$ that appears satisfies $s_1+\ldots+s_k=5$).

There is a ton of literature on producing homogeneous relations among multiple zeta values. One general class of relations, called the extended double shuffle relations, is conjectured to include all relations.

You identity follows by taking a linear combination of the following MZV relations, which can be found in the literature: the double shuffle relation applied to $\zeta(2)\zeta(3)$ (see Derivation and double shuffle relations for multiple zeta values, by Ihara, Kaneko, and Zagier): $$ \zeta(5)=2\zeta(3,2)+6\zeta(4,1), $$ duality applied to $\zeta(3,1,1)$ (this is a consequence of an expression of Konstsevich for MZV as iterated integrals, see The algebra of multiple harmonic series, by Hoffman): $$ \zeta(3,1,1)=\zeta(4,1), $$ and the sum formula (known in this case by Euler and proven in the general case independently by Granville and Zagier, the statement can also be found in The algebra of multiple harmonic series): $$ \zeta(2,3)+\zeta(3,2)+\zeta(4,1)=\zeta(5). $$


Identities such as this can be proved with the help of Cauchy's residue theorem. If $f$ is a meromorphic function such that $\lvert f(z)\rvert=o(z^{-1})$ as $\lvert z\rvert\to\infty$ on a sequence of concentric circles about the origin then, the residue theorem gives $$ \begin{align} \sum_a {\rm Res}(f,a)=0.&&{\rm(1)} \end{align} $$ Here, the sum is over all poles of $f$ and ${\rm Res}(f,a)$ is the residue of $f$ at $a$. The tricky part is finding the right function $f$. Flajolet & Salvy1 show how to prove a whole set of identities of this form. For example (all sums are over $n=1$ to $\infty$), $$ \begin{align} &\sum\frac{H_n}{n^2}=2\zeta(3),\\ &\sum\frac{H_n}{n^3}=\frac54\zeta(4),\\ &\sum\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3),\\ &\sum\frac{(H_n)^2}{n^2}=\frac{17}{4}\zeta(4),\\ &\sum\frac{(H_n)^2}{n^3}=\frac72\zeta(5)-\zeta(2)\zeta(3),&&{\rm(2)}\\ &\sum\frac{(H_n)^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta(3)^2,\\ &\sum\frac{(H_n)^3}{n^4}=\frac{231}{16}\zeta(7)-\frac{51}{4}\zeta(3)\zeta(4)+2\zeta(2)\zeta(5) \end{align} $$ We also have the set of identities due to Euler (for $q\ge2$) $$ \sum_{n=1}^\infty\frac{H_n}{n^q}=\left(1+\frac q2\right)\zeta(q+1)-\frac12\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k). $$ It is mentioned in Flajolet & Salvy that identities of this form do not always exist and, in particular, there is unlikely to be any finite formula for $\sum (H_n)^3/n^q$ in terms of zeta values when $q$ is an odd number exceeding $10$.

I'll give the function $f$ which generates the identity (2) asked for, following Flajolet & Salvy (specializing to this example). Let $\psi$ be the digamma function $$ \psi(z)=\frac{d}{dz}\log\Gamma(z)=-\gamma-\frac1z+\sum_{n=1}^\infty\left(\frac1n-\frac1{n+z}\right). $$ This is bounded by $O(\lvert z\rvert^\epsilon)$ on circles of radius $n+1/2$ about the origin and has poles at the nonnegative integers. To prove the required identity (2), it is easiest to break this down into three identities (although, you could add the three choices of $f$ below and do it in one go).

Taking the function $f(z)=\frac13z^{-3}\left(\psi(-z)+\gamma\right)^3$, this has poles at the nonnegative integers. Expand each term about $0$ and positive integers $n\gt0$, $$ \begin{align} &\psi(-z)+\gamma=\frac1z-\zeta(2)z-\zeta(3)z^2-\zeta(4)z^3-\zeta(5)z^4+O(z^5)\\ &\psi(-n-z)+\gamma=\frac1z+H_n-\left(H_n^{(2)}+\zeta(2)\right)z+O(z^2)\\ &(n+z)^{-3}=n^{-3}-3n^{-4}z+6n^{-5}z^2+O(z^3) \end{align} $$ Here, $H_n^{(2)}$ is the generalized harmonic number $\sum_{k\le n}k^{-2}$. Multiplying the terms together and extracting the coefficients of $z^{-1}$ gives the residues of $f$. $$ \begin{align} &{\rm Res}(f,0)=2\zeta(2)\zeta(3)-\zeta(5),\\ &{\rm Res}(f,n)=n^{-3}(H_n^2-H_n^{(2)}-\zeta(2))-3n^{-4}H_n+2n^{-5}. \end{align} $$ Summing over $n$ and applying the residue theorem, $$ \begin{align} \sum\frac{H_n^2}{n^3}-\sum\frac{H_n^{(2)}}{n^3}-3\sum\frac{H_n}{n^4}+\zeta(5)+\zeta(2)\zeta(3)=0.&&{\rm(3)} \end{align} $$ Now, take the function $f(z)=\frac12z^{-4}\left(\psi(-z)+\gamma\right)^2$. This again has poles at the nonnegative integers. Using the expansions above together with $$ (n+z)^{-4}=n^{-4}-4n^{-5}z+O(z^2) $$ we can compute the residues as before $$ \begin{align} &{\rm Res}(f,0)=\zeta(2)\zeta(3)-\zeta(5),\\ &{\rm Res}(f,n)=n^{-4}H_n-2n^{-5}. \end{align} $$ Applying the residue theorem again gives $$ \begin{align} \sum\frac{H_n}{n^4}-3\zeta(5)+\zeta(2)\zeta(3)=0.&&{\rm(4)} \end{align} $$ Finally take $f(z)=\frac12\pi z^{-3}\cot(\pi z)\psi^\prime(-z)$ and use the expansions $$ \begin{align} &\psi^\prime(-z)=z^{-2}+\zeta(2)+2\zeta(3)z+3\zeta(4)z^2+4\zeta(5)z^3+O(z^4)\\ &\psi^\prime(-n-z)=z^{-2}+H_n^{(2)}+\zeta(2)+O(z)\\ &\psi^\prime(n-z)=\zeta(2)+n^{-2}-H_n^{(2)}+O(z)\\ &\pi\cot(\pi(\pm n+z))=z^{-1}-\frac13\pi^2z+cz^3+O(z^5)=z^{-1}-2\zeta(2)z+cz^3+O(z^5) \end{align} $$ (some constant $c$) to compute the residues $$ \begin{align} &{\rm Res}(f,0)=2\zeta(5)-2\zeta(2)\zeta(3),\\ &{\rm Res}(f,n)=\frac12n^{-3}(H_n^{(2)}-\zeta(2))+3n^{-5},\\ &{\rm Res}(f,-n)=-\frac12n^{-3}(\zeta(2)+n^{-2}-H_n^{(2)}). \end{align} $$ Summing over $n$ and applying the residue theorem, $$ \begin{align} \sum\frac{H_n^{(2)}}{n^3}+\frac92\zeta(5)-3\zeta(2)\zeta(3)=0.&&{\rm(5)} \end{align} $$ Adding identities (3), 3 times (4), and (5) gives the required result.

I'll just add a note that the use of the digamma function, cotangent and residue theorem above are not really required. It has been mentioned in Noam D. Elkie's answer that such results can be proved by elementary, but clever, algebraic manipulations. Applied to rational functions, the residue theorem gives algebraic identities which can be easily verified. Also, the digamma function and cotangent can be expressed as sums over terms of the form $i^{-1}-(i+z)^{-1}$ over integer $i$. So, expanding the functions $f$ above as infinite sums over rational functions before applying the residue theorem reduces the argument to one involving summing over elementary identities. In particular, applying the residue theorem to the functions $\frac1{z^2}(\frac1i-\frac1{i-z})(\frac1j-\frac1{j-z})$, $\frac1{z^3}(\frac1i-\frac1{i+z})\frac1{(j-z)^2}$ and $\frac1{z^3}(\frac1i-\frac1{i-z})(\frac1j-\frac1{j-z})(\frac1k-\frac1{k-z})$ gives, respectively, $$ \begin{align} &\frac1{i^4}\left(\frac1j-\frac1{j-i}\right)+ \frac1{j^4}\left(\frac1i-\frac1{i-j}\right)+\frac1{i^2j^3}+\frac1{i^3j^2}=0,\\ &\frac3{j^4}\left(\frac1i-\frac1{i+j}\right)-\frac1{j^3(i+j)^2}-\frac1{i^3(i+j)^2}-\frac2{i^2j^3}+\frac1{i^3j^2}=0,\\ &\sum_{(ijk)}\frac1{i^3}\left(\frac1j-\frac1{j-i}\right)\left(\frac1k-\frac1{k-i}\right)=0. \end{align} $$ In the last identity, the summation refers to the sum over the three cyclic permutations of $i,j,k$. Summing these identities over positive integers $i,j,k$ and cancelling terms of the form $\frac1i$ and $\frac{-1}{i\pm j}$ leads to identities (3,4,5) above.

1 Euler sums and contour integral representations, P. Flajolet, B. Salvy, Experimental Mathematics Volume 7, Issue 1 (1998), 15-35. (link)


This is a known result; an explicit reference is [PP, equation (3c)]. The proof there is elementary (no need for digamma functions, definite integrals and contour integrals, etc.), but nontrivial, requiring clever manipulations with identities such as $1/XY = 1/(X(X+Y)) + 1/(Y(X+Y))$. As Julian Rosen noted, such sums are often expressed in terms of multiple zeta functions, such as the double and triple zetas $$ \zeta(a,b) = \mathop{\sum\sum}_{0<m<n} \frac1{m^a n^b}, \quad \zeta(a,b,c) = \mathop{\sum\sum\sum}_{0<l<m<n} \frac1{l^a m^b n^c}. $$ I found [PP] via Michael Hoffmann's list of References on multiple zeta values and Euler sums. [PP] cites a paper of Borwein and Girgensohn [BG], where the key triple-zeta value $\zeta(3,1,1) = 2 \zeta(5) - \zeta(3) \zeta(2)$ is given on page 21, together with the note that all such values of weight at most 6 appear in [M]. Indeed here the weight is $3+1+1 = 5 \leq 6$ and the result is the case $p=3$ of Theorem 4.1, listed explicitly as equation (4.2) on page 126.


[BG] J. Borwein and R. Girgensohn, Evaluation of triple Euler sums, Electronic Journal of Combinatorics 3, research paper #23, 1996.

[M] C. Markett: Triple sums and the Riemann zeta function, J. Number Theory 48 (1994), 113$-$132.

[PP] Alois Panholzer and Helmut Prodinger: Computer-free evaluation of an infinite double sum via Euler sums, Séminaire Lotharingien de Combinatoire 55 (2005), Article B55a.

  • $\begingroup$ Since you are referring to the paper of Panholzer and Prodinger, it may be of interest for you that there's a typo in the main formula. If you like you may have a look at my answer to this question. The formula you are referring here is correct. Regards. $\endgroup$ – Markus Scheuer Sep 10 '14 at 21:29
  • $\begingroup$ Thanks for this warning and the link. $\endgroup$ – Noam D. Elkies Sep 10 '14 at 23:13

Since the sum in question belongs to the general category of non-linear Euler sums we use the following notation: \begin{eqnarray} {\bf H}^{(p,q)}_r(t) &:=& \sum\limits_{m=1}^\infty H_m^{(p)} H_m^{(q)} \cdot \frac{t^m}{m^r}\\ {\bf H}^{(p)}_r(t) &:=& \sum\limits_{m=1}^\infty H_m^{(p)} \cdot \frac{t^m}{m^r}\\ \end{eqnarray} Now by using my answer to Double harmonic sum $\sum_{n\geq 1}\frac{H^{(p)}_nH_n}{n^q}$ I have generated the following results: \begin{eqnarray} {\bf H}^{(1,1)}_2(+1) &=& \frac{17 \zeta(4)}{4}\\ {\bf H}^{(1,1)}_3(+1) &=&\frac{7 \zeta (5)}{2}-\zeta (3) \zeta(2)\\ {\bf H}^{(1,1)}_4(+1) &=&\frac{97 \zeta(6)}{24}-2 \zeta (3)^2\\ {\bf H}^{(1,1)}_5(+1) &=&-\zeta (5) \zeta(2)-\frac{5}{2} \zeta (3) \zeta(4)+6 \zeta (7)\\ {\bf H}^{(1,1)}_6(+1) &=&{\bf H}^{(2)}_6(+1)+\zeta(3)^2 \zeta(2)-8 \zeta (3) \zeta (5)+\frac{91 \zeta(8)}{12}\\ {\bf H}^{(1,1)}_7(+1) &=&-\zeta (7) \zeta(2)-\frac{5}{2} \zeta (5) \zeta(4)-\frac{7}{2} \zeta (3) \zeta(6)+\frac{55 \zeta (9)}{6}+\frac{\zeta(3)^3}{3}\\ {\bf H}^{(1,1)}_8(+1) &=&{\bf H}^{(2)}_8(+1)+2 \zeta (3) \zeta (5) \zeta(2)+\zeta (3)^2 \zeta(4)-10 \zeta (3) \zeta (7)-5 \zeta (5)^2+\frac{473 \zeta(10)}{40}\\ {\bf H}^{(1,1)}_9(+1) &=&-\zeta (9) \zeta(2)-\frac{5}{2} \zeta (7)\zeta(4)-\frac{7}{2} \zeta (5) \zeta(6)-\frac{9}{2} \zeta (3) \zeta(8)+13 \zeta (11)+\zeta (3)^2 \zeta (5)\\ {\bf H}^{(1,1)}_{10}(+1) &=&{\bf H}^{(2)}_{10}(+1)+2 \zeta (3) \zeta (7) \zeta(2)+\zeta (5)^2 \zeta(2)+2 \zeta (3) \zeta (5) \zeta(4)+\zeta (3)^2 \zeta(6)-12 \zeta (3) \zeta (9)-12 \zeta (5) \zeta (7)+\frac{23595 \zeta(12)}{1382}\\ {\bf H}^{(1,1)}_{11}(+1) &=& -\zeta (11) \zeta(2)-\frac{5}{2} \zeta (9) \zeta(4)-\frac{7}{2} \zeta (7) \zeta(6)-\frac{9}{2} \zeta (5) \zeta(8)-\frac{11}{2} \zeta (3) \zeta(10)+\frac{35 \zeta (13)}{2}+\zeta (3)^2 \zeta (7)+\zeta (3) \zeta (5)^2\\ {\bf H}^{(1,1)}_{12}(+1) &=& {\bf H}^{(2)}_{12}(+1)+2 \zeta (3) \zeta (9) \zeta(2)+2 \zeta (5) \zeta (7) \zeta(2)+2 \zeta (3) \zeta (7) \zeta(4)+\zeta (5)^2 \zeta(4)+2 \zeta (3) \zeta (5) \zeta(6)+\zeta (3)^2 \zeta(8)-14 \zeta (3) \zeta (11)-14 \zeta (5) \zeta (9)-7 \zeta (7)^2+\frac{19591 \zeta(14)}{840}\\ {\bf H}^{(1,1)}_{13}(+1) &=& -\zeta (13) \zeta(2)-\frac{5}{2} \zeta (11) \zeta(4)-\frac{7}{2} \zeta (9) \zeta(6)-\frac{9}{2} \zeta (7) \zeta(8)-\frac{11}{2} \zeta (5) \zeta(10)-\frac{13}{2} \zeta (3) \zeta(12)+\frac{68 \zeta (15)}{3}+\zeta (3)^2 \zeta (9)+2 \zeta (3) \zeta (5) \zeta (7)+\frac{\zeta (5)^3}{3}\\ {\bf H}^{(1,1)}_{14}(+1) &=& {\bf H}^{(2)}_{14}(+1)+2 \zeta (3) \zeta (11) \zeta(2)+2 \zeta (5) \zeta (9) \zeta(2)+\zeta (7)^2 \zeta(2)+2 \zeta (3) \zeta (9) \zeta(4)+2 \zeta (5) \zeta (7) \zeta(4)+2 \zeta (3) \zeta (7) \zeta(6)+\zeta (5)^2 \zeta(6)+2 \zeta (3) \zeta (5) \zeta(8)+\zeta (3)^2 \zeta(10)-16 \zeta (3) \zeta (13)-16 \zeta (5) \zeta (11)-16 \zeta (7) \zeta (9)+\frac{442323 \zeta(16)}{14468} \end{eqnarray} Note that whenever $r$ is odd then the result reduces to single zeta function values only. On the other hand if $r$ is even then one additional parameter appears.

  • $\begingroup$ @ Przemo Well done. Did you also consider alternating sums ($t=-1$)? $\endgroup$ – Dr. Wolfgang Hintze Jun 6 '19 at 13:08

using the following identity : $$\displaystyle \frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)x^n$$

multiply both sides by $\frac{\ln^2x}{x}$ then integrate both sides w.r.t $x$ from $0$ to $1$, we have: \begin{align*} S&=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)\int_0^1x^{n-1}\ln^2x\ dx=\color{blue}{2\sum_{n=1}^{\infty}\frac{H_n^2-H_n^{(2)}}{n^3}}=\int_0^1\frac{\ln^2x\ln^2(1-x)}{x(1-x)}\ dx\\ &=\int_0^1\frac{\ln^2x\ln^2(1-x)}{x}\ dx+\underbrace{\int_0^1\frac{\ln^2x\ln^2(1-x)}{1-x}\ dx}_{x\mapsto1-x}=2\int_0^1\frac{\ln^2x\ln^2(1-x)}{x}\ dx\\ &=4\sum_{n=1}^{\infty}\left(\frac{H_n}{n}-\frac1{n^2}\right)\int_0^1x^{n-1}\ln^2x\ dx=4\sum_{n=1}^{\infty}\left(\frac{H_n}{n}-\frac1{n^2}\right)\frac{2}{n^3}=\color{blue}{8\sum_{n=1}^{\infty}\frac{H_n}{n^4}-8\zeta(5)} \end{align*}

where we used $\ln^2(1-x)=2\sum_{n=1}^\infty\frac{H_n}{n+1}x^{n+1}=2\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac1{n^2}\right)x^n$

rearranging the terms of the blue sides, we have: \begin{align*} \sum_{n=1}^{\infty}\frac{H_n^2}{n^3}&=\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^3}+4\sum_{n=1}^{\infty}\frac{H_n}{n^4}-4\zeta(5)\\ &=\left(3\zeta(2)\zeta(3)-\frac92\zeta(5)\right)+4\left(3\zeta(5)-\zeta(2)\zeta(3)\right)-4\zeta(5)\\ &=\frac72\zeta(5)-\zeta(2)\zeta(3) \end{align*}

The proof of $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$ can be found here.


Since $$\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n=\frac12\ln^2(1-x)$$

divide both sides by $x$ then $\int_0^y$


Next multiply both sides by $-\frac{\ln(1-y)}{y}$ then $\int_0^1$ and apply that $-\int_0^1 y^{n-1}\ln(1-y)dy=\frac{H_n}{n}$





$$=\zeta(2)\zeta(3)-\frac12\sum_{n=1}^\infty\frac1{n^2}\int_0^1 x^{n-1}\ln^2(1-x)dx$$



Rearrange the terms


where we used $\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ and $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$


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