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.
Real Analysis – Infinite Series Sum of (H_n)^2/n^3
closed-formcomplex-analysisharmonic-numbersreal-analysissequences-and-series
Related Solutions
EDITED. Some simplifications were made.
Here is a solution.
1. Basic facts on the dilogarithm. Let $\mathrm{Li}_{2}(z)$ be the dilogarithm function defined by
$$ \operatorname{Li}_{2}(z) = \sum_{n=1}^{\infty} \frac{z^{n}}{n^{2}} = - \int_{0}^{z} \frac{\log(1-x)}{x} \, dx. $$
Here the branch cut of $\log $ is chosen to be $(-\infty, 0]$ so that $\operatorname{Li}_{2}$ defines a holomorphic function on the region $\Bbb{C} \setminus [1, \infty)$. Also, it is easy to check (by differentiating both sides) that the following identities hold
\begin{align*} \operatorname{Li}_{2}\left(\tfrac{z}{z-1}\right) &= -\mathrm{Li}_{2}(z) - \tfrac{1}{2}\log^{2}(1-z); \quad z \notin [1, \infty) \tag{1} \\ \operatorname{Li}_{2}\left(\tfrac{1}{1-z}\right) &= \color{blue}{\boxed{\operatorname{Li}_{2}(z) + \zeta(2) - \tfrac{1}{2}\log^{2}(1-z)}} + \color{red}{\boxed{\log(-z)\log(1-z)}}; \quad z \notin [0, \infty) \tag{2} \end{align*}
Notice that in (2), the blue-colored part is holomorphic on $|z| < 1$ while the red-colored part induces the branch cut $[-1, 0]$.
2. A useful power series. Now let us consider the power series
$$ f(z) = \sum_{n=0}^{\infty} \frac{H_n}{n} z^n. $$
Then $f(z)$ is automatically holomorphic inside the disc $|z| < 1$. Moreover, it is easy to check that
$$ \sum_{n=1}^{\infty} H_{n} z^{n-1} = \frac{1}{z} \left( \sum_{n=1}^{\infty} \frac{z^{n}}{n} \right)\left( \sum_{n=0}^{\infty} z^{n}\right) = -\frac{\log(1-z)}{z(1-z)}. $$
thus integrating both sides, together with the identity $\text{(1)}$, we obtain the following representation of $f(z)$.
$$f(z) = \operatorname{Li}_{2}(z) + \tfrac{1}{2}\log^{2}(1-z) = -\operatorname{Li}_{2}\left(\tfrac{z}{z-1}\right). \tag{3}$$
3. Integral representation and the result. By the Parseval's identity, we have
$$ \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}} = \frac{1}{2\pi} \int_{0}^{2\pi} f(e^{it})f(e^{-it}) \, dt = \frac{1}{2\pi i} \int_{|z|=1} \frac{f(z)}{z} f\left(\frac{1}{z}\right) \, dz \tag{4} $$
Since $\frac{1}{z}f(z)$ is holomorphic inside $|z| = 1$, the failure of holomorphy of the integrand stems from the branch cut of
\begin{align*} f\left(\tfrac{1}{z}\right) &= -\operatorname{Li}_{2}\left(\tfrac{1}{1-z}\right) \\ &= -\color{blue}{\left( \operatorname{Li}_{2}(z) + \zeta(2) - \tfrac{1}{2}\log^{2}(1-z) \right)} - \color{red}{\log(-z)\log(1-z)}, \end{align*}
which is $[0, 1]$. To resolve this, we utilize the identity $\text{(2)}$. Note that the blue-colored portion does not contributes to the the integral $\text{(4)}$, since it remains holomorphic inside $|z| < 1$. That is, only the red-colored portion gives contribution to the integral. Consequently we have
\begin{align*} \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}} &= -\frac{1}{2\pi i} \int_{|z|=1} \frac{f(z)}{z} \color{red}{\log(-z)\log(1-z)} \, dz. \tag{5} \end{align*}
Since the integrand is holomorphic on $\Bbb{C} \setminus [0, \infty)$, we can utilize the keyhole contour wrapping around $[0, 1]$ to reduce $\text{(5)}$ to
\begin{align*} \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}} &=-\frac{1}{2\pi i} \Bigg\{ \int_{0^{-}i}^{1+0^{-}i} \frac{f(z)\log(-z)\log(1-z)}{z} \, dz \\ &\qquad \qquad + \int_{1+0^{+}i}^{+0^{+}i} \frac{f(z)\log(-z)\log(1-z)}{z} \, dz \Bigg\} \\ &=-\frac{1}{2\pi i} \Bigg\{ \int_{0}^{1} \frac{f(x)(\log x + i\pi)\log(1-x)}{x} \, dx \\ &\qquad \qquad - \int_{0}^{1} \frac{f(x)(\log x - i\pi)\log(1-x)}{x} \, dx \Bigg\} \\ &=-\int_{0}^{1} \frac{f(x)\log(1-x)}{x} \, dx. \tag{5} \end{align*}
Plugging $\text{(3)}$ to the last integral and simplifying a little bit, we have
\begin{align*} \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}} &= - \int_{0}^{1} \frac{\operatorname{Li}_2(x)\log(1-x)}{x} \, dx - \frac{1}{2}\int_{0}^{1} \frac{\log^{3}(1-x)}{x} \, dx \\ &= \left[ \frac{1}{2}\operatorname{Li}_2(x)^2 \right]_0^1 - \frac{1}{2} \int_{0}^{1} \frac{\log^3 x}{1-x} \, dx \\ &= \frac{1}{2}\zeta(2)^{2} + \frac{1}{2} \Gamma(4)\zeta(4) \\ &= \frac{17\pi^{4}}{360} \end{align*}
as desired.
Let's start with the product of $\;-\ln(1-x)\,$ and $\dfrac 1{1-x}$ to get the product generating function
(for $|x|<1$) :
$$\tag{1}f(x):=-\frac {\ln(1-x)}{1-x}=\sum_{n=1}^\infty H_n\, x^n$$
Dividing by $x$ and integrating we get :
\begin{align}
\sum_{n=1}^\infty \frac{H_n}n\, x^n&=\int \frac{f(x)}xdx\\
&=-\int \frac{\ln(1-x)}{1-x}dx-\int\frac{\ln(1-x)}xdx\\
\tag{2}&=C+\frac 12\ln(1-x)^2+\operatorname{Li}_2(x)\\
\end{align}
(with $C=0$ from $x=0$)
The first integral was obtained by integration by parts, the second from the integral definition of the dilogarithm or the recurrence for the polylogarihm (with $\;\operatorname{Li}_1(x)=-\ln(1-x)$) : $$\tag{3}\operatorname{Li}_{s+1}(x)=\int\frac {\operatorname{Li}_{s}(x)}x dx$$
Dividing $(2)$ by $x$ and integrating again returns (using $(3)$ again) :
\begin{align}
\sum_{n=1}^\infty \frac{H_n}{n^2}\, x^n&=\int \frac {\ln(1-x)^2}{2\,x}dx+\int \frac{\operatorname{Li}_2(x)}x dx\\
&=C+I(x)+\operatorname{Li}_3(x)\\
\end{align}
with $I(x)$ obtained by integration by parts (since $\frac d{dx}\operatorname{Li}_2(1-x)=\dfrac {\ln(x)}{1-x}$) :
\begin{align}
I(x)&:=\int \frac {\ln(1-x)^2}{2\,x}dx\\
&=\left.\frac{\ln(1-x)^2\ln(x)}{2}\right|+\int \ln(1-x)\frac {\ln(x)}{1-x}dx\\
&=\left.\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)\right|+\int \frac{\operatorname{Li}_2(1-x)}{1-x}dx\\
&=\left.\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)-\operatorname{Li}_3(1-x)\right|\\
\end{align}
getting the general relation :
$$\tag{4}\sum_{n=1}^\infty \frac{H_n}{n^2}\, x^n=C+\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)$$
(with $C=\operatorname{Li}_3(1)=\zeta(3)$ here)
applied to $x=\dfrac 12$ with $\operatorname{Li}_2\left(\frac 12\right)=\dfrac{\zeta(2)-\ln(2)^2}2$ from the link returns the wished :
\begin{align}
\sum_{n=1}^\infty \frac{H_n}{n^2\;2^n}&=\zeta(3)-\frac{\ln(2)^3}2-\ln(2)\frac{\zeta(2)-\ln(2)^2}2\\
\tag{5}\sum_{n=1}^\infty \frac{H_n}{n^2\;2^n}&=\zeta(3)-\ln(2)\frac{\zeta(2)}2
\end{align}
Best Answer
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)