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this series was evaluated by Cornel Valean here using series manipulation.

I took a different path as follows:

using the identity:$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\right)$$ multiply both sides by $\ln^3x/x$ then integrate

$$-6\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n^4}=\int_0^1\frac{\ln^2(1-x)\ln^3x}{x(1-x)}\ dx$$ I was able here to find \begin{align} \sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}&=\frac43\zeta^2(3)-\frac23\sum_{k=1}^\infty\frac{H_k^{(3)}}{k^3}\\ &=\zeta^2(3)-\frac13\zeta(6) \end{align} as for the integral, it seems very tedious to calculate it using the derivative of beta function.

can we find it with or without using beta function?

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  • $\begingroup$ By 'as for the integral', do you mean $\int_0^1\frac{\ln^2(1-x)\ln^3x}{x(1-x)}\ dx$ which equals to $[\frac{\ln^6 x}{6}]_0^1$? $\endgroup$
    – Toby Mak
    Jun 2, 2019 at 6:08
  • $\begingroup$ @TobyMak how's that? That integral is not easy to calculate using beta function. $\endgroup$ Jun 2, 2019 at 6:36
  • $\begingroup$ You cannot compute the integral since at $x=0$ the integral does not converge. $\endgroup$
    – Toby Mak
    Jun 2, 2019 at 12:03
  • $\begingroup$ Using the derivative of the Beta function seems to be the clearer way. Anyway have you tried to split into two integrals with partial fraction? One has: $$\int_0^1 \frac{\ln^2(1-x)\ln^3 x}{x}dx=-12\sum_{n=1}^\infty \frac{H_n}{(n+1)^5}$$ $\endgroup$
    – Zacky
    Jun 2, 2019 at 15:24
  • $\begingroup$ @TobyMak using the limit as x approaches zero is not a problem and the integral converges. $\endgroup$ Jun 2, 2019 at 15:37

2 Answers 2

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This solution is by Cornel Valean.

Using the follwing identity: ( see Lemma $2(b)$ in this paper) $$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$ and since $$\int_0^1x^{n-1}\ln^2(1-x)\ dx=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\int_0^1x^{n+k-1}\ dx=2\sum_{k=1}^\infty\frac{H_{k-1}}{k(n+k)}$$ Then $$\sum_{k=1}^\infty\frac{H_{k-1}}{k(n+k)}=\frac{H_n^2+H_n^{(2)}}{2n}\tag{1}$$ Divide both sides by $n^3$ then sum both sides from $n=1$ to $\infty$, we get \begin{align} S&=\color{blue}{\frac12\sum_{n=1}^\infty\frac{H_n^2}{n^4}+\frac12\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}}=\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\sum_{n=1}^\infty\frac{1}{n^3(n+k)}\right)\\ &=\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\sum_{n=1}^\infty\left[\frac{1}{k^3}\left(\frac{1}{n}-\frac{1}{n+k}\right)-\frac1{n^2k^2}+\frac1{n^3k}\right]\right)\\ &=\sum_{k=1}^\infty\left(\frac{H_k}{k}-\frac{1}{k^2}\right)\left(\frac{H_k}{k^3}-\frac{\zeta(2)}{k^2}+\frac{\zeta(3)}{k}\right)\\ &=\sum_{k=1}^\infty\frac{H_k^2}{k^4}-\sum_{k=1}^\infty\frac{H_k}{k^5}-\zeta(2)\sum_{k=1}^\infty\left(\frac{H_k}{k^3}-\frac1{k^4}\right)+\zeta(3)\sum_{k=1}^\infty\left(\frac{H_k}{k^2}-\frac1{k^3}\right)\\ &=\sum_{k=1}^\infty\frac{H_k^2}{k^4}-\left(\frac74\zeta(6)-\frac12\zeta^2(3)\right)-\zeta(2)\left(\frac14\zeta(4)\right)+\zeta(3)\left(\zeta(3)\right)\\ &=\color{blue}{\sum_{k=1}^\infty\frac{H_k^2}{k^4}-\frac{35}{16}\zeta(6)+\frac32\zeta^2(3)} \end{align} Rearranging the blue sides, we get

$$\sum_{k=1}^\infty\frac{H_k^2}{k^4}=\frac{35}{8}\zeta(6)-3\zeta^2(3)+\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}\\ =\frac{97}{24}\zeta(6)-2\zeta^2(3)$$

where we used $\ \displaystyle\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}=\zeta^2(3)-\frac13\zeta(6)\ $ (can be found in the same paper I linked or here)

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For a slight variation on a theme.

As $$\int_0^1 x^{n - 1} \ln^2 (1 - x) \, dx = \frac{H^2_n}{n} + \frac{H^{(2)}_n}{n},$$ for a proof of this result, see here, we can write the sum as \begin{align} \sum_{n = 1}^\infty \frac{H^2_n}{n^4} &= \sum_{n = 1}^\infty \frac{1}{n^3} \cdot \frac{H^2_n}{n}\\ &= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + \int_0^1 \frac{\ln^2 (1 - x)}{x} \sum_{n = 1}^\infty \frac{x^n}{n^3} \, dx\\ &= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + \int_0^1 \frac{\ln^2 (1 - x) \operatorname{Li}_3 (x)}{x} \, dx.\tag1 \end{align} Making use of the following Maclaurin series expansion for $\ln^2 (1 - x)$, namely $$\ln^2 (1 - x) = 2 \sum_{n = 1}^\infty \frac{H_n x^{n + 1}}{n + 1},$$ the integral in (1) can be re-written as \begin{align} \sum_{n = 1}^\infty \frac{H^2_n}{n^4} &= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \sum_{n = 1}^\infty \frac{H_n}{n + 1} \underbrace{\int_0^1 x^n \operatorname{Li}_3 (x) \, dx}_{\text{IBP 3 times}}\\ &= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \sum_{n = 1}^\infty \frac{H_n}{n+ 1} \left [\frac{\zeta (3)}{n + 1} - \frac{\zeta (2)}{(n + 1)^2} + \frac{H_{n + 1}}{(n + 1)^3} \right ]\\ &= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \zeta (3) \underbrace{\sum_{n = 1}^\infty \frac{H_n}{(n + 1)^2}}_{n \, \mapsto \, n - 1} -2 \zeta (2) \underbrace{\sum_{n = 1}^\infty \frac{H_n}{(n + 1)^3}}_{n \, \mapsto \, n - 1} + 2 \underbrace{\sum_{n = 1}^\infty \frac{H_n H_{n + 1}}{(n + 1)^2}}_{n \, \mapsto \, n - 1}\\ &= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \zeta (3) \sum_{n = 1}^\infty \frac{1}{n^2} \left (H_n - \frac{1}{n} \right ) - 2 \zeta (2) \sum_{n = 1}^\infty \frac{1}{n^3} \left (H_n - \frac{1}{n} \right )\\ & \qquad + 2 \sum_{n = 1}^\infty \frac{H_n}{n^4} \left (H_n - \frac{1}{n} \right )\\ &= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \zeta (3) \sum_{n = 1}^\infty \frac{H_n}{n^2} - 2 \zeta^2 (3) - 2 \zeta (2) \sum_{n = 1}^\infty \frac{H_n}{n^3} + 2 \zeta (2) \zeta (4)\\ & \qquad + 2 \sum_{n = 1}^\infty \frac{H^2_n}{n^4} - 2 \sum_{n = 1}^\infty \frac{H_n}{n^5}\\ \Rightarrow \sum_{n = 1}^\infty \frac{H^2_n}{n^4} &= \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} - 2 \zeta (3) \sum_{n = 1}^\infty \frac{H_n}{n^2} + 2 \zeta (2) \sum_{n = 1}^\infty \frac{H_n}{n^3} + 2 \sum_{n = 1}^\infty \frac{H_n}{n^5}\\ & \qquad + 2 \zeta^2 (3) - 2 \zeta (2) \zeta (4).\tag2 \end{align}

Making use of the following results: \begin{align} \sum_{n = 1}^\infty \frac{H_n}{n^2} &= 2 \zeta (3)\\ \sum_{n = 1}^\infty \frac{H_n}{n^3} &= \frac{5}{4} \zeta (4)\\ \sum_{n = 1}^\infty \frac{H_n}{n^5} &= -\frac{1}{2} \zeta^2 (3) + \frac{7}{4} \zeta (6)\\ \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} &= \zeta^2 (3) - \frac{1}{3} \zeta (6)\\ \zeta (2) \zeta (4) &= \frac{7}{6} \zeta (6) \end{align} substituting into (2) leads to $$\sum_{n = 1}^\infty \frac{H^2_n}{n^4} = \frac{97}{24} \zeta (6) - 2 \zeta^2 (3),$$ as desired.

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  • $\begingroup$ Nice... I missed this approach. $\endgroup$ Jul 17, 2019 at 4:11
  • $\begingroup$ Its more convenient to write $$\ln^2(1-x)=2\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac{1}{n^2}\right)x^n$$ and this way we can avoid the annoying expression of $n+1$. $\endgroup$ Jul 17, 2019 at 4:24
  • $\begingroup$ @Ali Shather - Yes, I agree writing $\ln^2 (1 - x)$ that way would be more convenient. Nice observation. $\endgroup$
    – omegadot
    Jul 17, 2019 at 4:27
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    $\begingroup$ I always write it that way if you notice my solutions. by the way, we can apply the same two approaches to find $\sum_{n=1}^\infty \frac{H_n^2}{n^5}$ as its easy to compute $\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^5}=5\zeta(2)\zeta(5)+2\zeta(3)\zeta(4)-10\zeta(7)$ $\endgroup$ Jul 17, 2019 at 4:30

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