Very recently Cornel discovered two (update: in fact there are more as seen from the new entires) fascinating results involving harmonic series using ideas from his book, (Almost) Impossible Integrals, Sums, and Series, and which are the core of a new paper he's preparing:

\begin{equation*} \sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n)^3} \end{equation*} \begin{equation*} =\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) \end{equation*} \begin{equation*} -2 \log (2) \operatorname{Li}_4\left(\frac{1}{2}\right) -2 \operatorname{Li}_5\left(\frac{1}{2}\right); \end{equation*} and \begin{equation*} \sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n-1)^3} \end{equation*} \begin{equation*} =6 \log (2)-2 \log ^2(2)-\frac{1}{12}\log ^4(2)+\frac{1}{12} \log ^5(2)-\frac{3}{2} \zeta (2)-\frac{21}{8} \zeta (3)+\frac{173}{32} \zeta (4) \end{equation*} \begin{equation*} +\frac{527}{128} \zeta (5)-\frac{21 }{16}\zeta (2) \zeta (3)+\frac{3}{2} \log (2) \zeta (2)-\frac{7}{2}\log (2)\zeta (3)-4\log (2)\zeta (4)+\frac{1}{2} \log ^2(2) \zeta (2) \end{equation*} \begin{equation*} -\frac{1}{2} \log ^3(2)\zeta (2)+\frac{7}{4}\log ^2(2)\zeta (3)-2 \operatorname{Li}_4\left(\frac{1}{2}\right)+2 \log (2)\operatorname{Li}_4\left(\frac{1}{2}\right), \end{equation*} or, after adjustments, the form $$\sum _{n=1}^{\infty}\frac{H_n H_{2 n}}{(2 n+1)^3}$$ $$=\frac{1}{12}\log ^5(2)+\frac{31}{128} \zeta (5)-\frac{1}{2} \log ^3(2)\zeta (2)+\frac{7}{4} \log ^2(2) \zeta (3)-\frac{17}{8} \log (2)\zeta (4) \\+2\log (2) \operatorname{Li}_4\left(\frac{1}{2}\right).$$ Update I : A new series entry obtained based on the aforementioned series \begin{equation*} \sum _{n=1}^{\infty } \frac{H_n H_{2 n}^{(2)}}{(2 n)^2} \end{equation*} \begin{equation*} =\frac{23 }{32}\zeta (2) \zeta (3)-\frac{581}{128} \zeta (5)-\frac{2}{3}\log ^3(2) \zeta (2)+\frac{7}{4} \log^2(2)\zeta (3) +\frac{2}{15} \log ^5(2) \end{equation*} \begin{equation*} +4\log (2) \operatorname{Li}_4\left(\frac{1}{2}\right) +4 \operatorname{Li}_5\left(\frac{1}{2}\right); \end{equation*} Update II : Another new series entry obtained based on the aforementioned series \begin{equation*} \sum _{n=1}^{\infty } \frac{H_n H_{2 n}^2}{(2 n)^2} \end{equation*} \begin{equation*} =\frac{23 }{32}\zeta (2) \zeta (3)+\frac{917 }{128}\zeta (5)+\frac{2}{3} \log ^3(2)\zeta (2)-\frac{7}{4} \log ^2(2)\zeta (3)-\frac{2}{15} \log ^5(2) \end{equation*} \begin{equation*} -4 \log (2)\operatorname{Li}_4\left(\frac{1}{2}\right)-4 \operatorname{Li}_5\left(\frac{1}{2}\right); \end{equation*} Update III : And a new series entry from the same class of series with an unexpected (and outstanding) closed-form \begin{equation*} \sum _{n=1}^{\infty } \frac{H_{2n} H_{n}^{(2)}}{(2 n)^2}=\frac{101 }{64}\zeta (5)-\frac{5 }{16}\zeta (2) \zeta (3); \end{equation*} It's interesting to note that $\displaystyle \sum _{n=1}^{\infty } \frac{H_{n} H_{n}^{(2)}}{n^2}=\zeta(2)\zeta(3)+\zeta(5)$, which may be found calculated in the book, (Almost) Impossible Integrals, Sums, and Series, by series manipulations.

A note: The series from UPDATE III seems to be known in literature, and it already appeared here https://math.stackexchange.com/q/1868355 (see $(3)$).

Update IV : Again a new series entry from the same class of series \begin{equation*} \sum _{n=1}^{\infty } \frac{H_n^2 H_{2 n}}{(2 n)^2} \end{equation*} \begin{equation*} =\frac{9 }{16}\zeta (2) \zeta (3)+\frac{421 }{64}\zeta (5)+\frac{2}{3} \log ^3(2)\zeta (2) -\frac{7}{4} \log ^2(2)\zeta (3) -\frac{2}{15} \log^5(2) \end{equation*} \begin{equation*} -4 \log(2)\operatorname{Li}_4\left(\frac{1}{2}\right) -4 \operatorname{Li}_5\left(\frac{1}{2}\right); \end{equation*} Update V : A strong series - September 26, 2019 $$\sum _{n=1}^{\infty } \frac{H_{2n} H_n^{(2)}}{(2 n+1)^2}$$ $$=\frac{4}{3}\log ^3(2)\zeta (2) -\frac{7}{2}\log^2(2)\zeta (3)-\frac{21}{16}\zeta(2)\zeta(3)+\frac{713}{64} \zeta (5)-\frac{4}{15} \log ^5(2)$$ $$-8 \log (2)\operatorname{Li}_4\left(\frac{1}{2}\right) -8\operatorname{Li}_5\left(\frac{1}{2}\right);$$ Update VI : Three very challenging series - September 28, 2019 $$i) \ \sum _{n=1}^{\infty } \frac{H_n H_{2 n}^{(2)}}{(2 n+1)^2}$$ $$=\frac{35}{32} \zeta (2) \zeta (3)-\frac{651}{128} \zeta (5)+\frac{1}{3}\log^3(2)\zeta(2)-\frac{7}{4}\log^2(2)\zeta (3)+\frac{53}{16} \log (2)\zeta (4) -\frac{1}{30} \log ^5(2)$$ $$+4 \operatorname{Li}_5\left(\frac{1}{2}\right);$$ $$ii) \ \sum _{n=1}^{\infty } \frac{H_n H_{2 n}^2}{(2 n+1)^2}$$ $$=\frac{35}{32} \zeta (2) \zeta (3)+\frac{465}{128} \zeta (5)+\frac{1}{2}\log^3(2)\zeta(2)-\frac{7}{4}\log^2(2)\zeta (3)-\frac{11}{16} \log (2)\zeta (4) -\frac{1}{12} \log ^5(2)$$ $$-2\log(2) \operatorname{Li}_4\left(\frac{1}{2}\right);$$ $$iii) \ \sum _{n=1}^{\infty } \frac{H_n^2 H_{2 n}}{(2 n+1)^2}$$ $$=\frac{21}{16} \zeta (2) \zeta (3)-\frac{217}{64} \zeta (5)+\frac{2}{3}\log^3(2)\zeta(2)-\frac{7}{4}\log^2(2)\zeta (3)+ \log (2)\zeta (4) -\frac{1}{15} \log ^5(2)$$ $$+8\operatorname{Li}_5\left(\frac{1}{2}\right);$$ Update VII : Critical series relation used in the Update VI - September 28, 2019 $$i) \sum _{n=1}^{\infty } \frac{H_n H_{2 n}^2}{(2 n+1)^2}-\sum _{n=1}^{\infty } \frac{H_n H_{2 n}^{(2)}}{(2 n+1)^2}$$ $$=\frac{1}{6}\log ^3(2)\zeta (2) -4\log (2)\zeta (4)+\frac{279}{32} \zeta (5)-\frac{1}{20} \log ^5(2)-2 \log (2)\operatorname{Li}_4\left(\frac{1}{2}\right) -4 \operatorname{Li}_5\left(\frac{1}{2}\right);$$ $$ii) \ 4 \sum _{n=1}^{\infty } \frac{H_n H_{2 n}^2}{(2 n+1)^2}-\sum _{n=1}^{\infty } \frac{H_n^2 H_{2 n}}{(2 n+1)^2}$$ $$=\frac{49}{16} \zeta (2) \zeta (3)+\frac{1147}{64}\zeta (5)+\frac{4}{3}\log^3(2)\zeta (2) -\frac{21}{4} \log ^2(2)\zeta (3) -\frac{15}{4}\log (2)\zeta (4)-\frac{4}{15} \log ^5(2)$$ $$-8 \log (2)\operatorname{Li}_4\left(\frac{1}{2}\right) -8\operatorname{Li}_5\left(\frac{1}{2}\right),$$ where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}, \ m\ge1,$ designates the $n$th generalized harmonic number of order $m$, $\zeta$ represents the Riemann zeta function, and $\operatorname{Li}_n$ denotes the Polylogarithm function.

A note: for example, for those interested, one of the possible ways of calculating both series from UPDATE III and UPDATE IV is based on building a system of relations with the two series by exploiting $\displaystyle \int_0^1 x^{n-1} \log^2(1-x)\textrm{d}x=\frac{H_n^2+H_n^{(2)}}{n}$ and $\displaystyle \sum_{n=1}^{\infty} x^n(H_n^2-H_n^{(2)})=\frac{\log^2(1-x)}{1-x}$. Apart from this, the series from UPDATE III allows at least a (very) elegant approach by using different means.

Using the first series we may obtain (based on the series representation of $\log(1-x)\log(1+x)$ and the integral $\int_0^1 x^{n-1}\operatorname{Li}_2(x)\textrm{d}x$) a way for proving that $$\int_0^1 \frac{\operatorname{Li}_2(x) \log (1+x) \log (1-x)}{x} \textrm{d}x=\frac{29 }{64}\zeta (5)-\frac{5 }{8}\zeta (2) \zeta (3).$$

Then, based on the solution below and using the alternating harmonic series in the book, (Almost) Impossible Integrals, Sums, and Series, we have
$$\int_0^1 \frac{\operatorname{Li}_2(-x) \log (1+x) \log (1-x)}{x} \textrm{d}x$$ $$=\frac{5 }{16}\zeta (2) \zeta (3)+\frac{123 }{32}\zeta (5)+\frac{2}{3} \log ^3(2)\zeta (2)-\frac{7}{4} \log ^2(2)\zeta (3)-\frac{2}{15}\log ^5(2)\\-4 \log (2)\operatorname{Li}_4\left(\frac{1}{2}\right)-4 \operatorname{Li}_5\left(\frac{1}{2}\right).$$ And if we add up the two previous integrals, we get $$\int_0^1 \frac{\operatorname{Li}_2(x^2) \log (1+x) \log (1-x)}{x} \textrm{d}x$$ $$=\frac{275}{32}\zeta (5)-\frac{5 }{8}\zeta (2) \zeta (3)+\frac{4}{3} \log ^3(2)\zeta (2)-\frac{7}{2} \log ^2(2)\zeta (3)-\frac{4}{15}\log ^5(2)\\-8 \log (2)\operatorname{Li}_4\left(\frac{1}{2}\right)-8 \operatorname{Li}_5\left(\frac{1}{2}\right).$$ Update (integrals): Another curious integral arising during the calculations $$\int_0^1 \frac{x \log (x) \log(1-x^2) \operatorname{Li}_2(x)}{1-x^2} \textrm{d}x=\frac{41 }{32}\zeta (2) \zeta (3)-\frac{269 }{128}\zeta (5).$$

QUESTION: Have these series ever been known in literature? I'm not interested in solutions but only if the series appear anywhere in the literature.

  • 2
    $\begingroup$ David Bailey would be as likely to know as anyone. $\endgroup$ Sep 5 '19 at 12:29
  • 1
    $\begingroup$ Cornel is the real boss. $\endgroup$ Sep 5 '19 at 17:09
  • $\begingroup$ Have you tried to contact Bailey, 97357329? $\endgroup$ Sep 6 '19 at 13:09
  • $\begingroup$ @GerryMyerson May I guess you refer to David H. Bailey? en.wikipedia.org/wiki/David_H._Bailey_(mathematician). I didn't yet. I don't know him personally. $\endgroup$ Sep 6 '19 at 13:13
  • 1
    $\begingroup$ Yes, that's the man. He's very approachable & I think he'd be interested. $\endgroup$ Sep 6 '19 at 22:50

Update: the paper mentioned below is the preprint On the calculation of two essential harmonic series with a weight $5$ structure, involving harmonic numbers of the type $H_{2n}$ by Cornel Ioan Valean

A solution in large steps by Cornel Ioan Valean:

Considering $\displaystyle -\log(1+y)\log(1-y)=\sum_{n=1}^{\infty} y^{2n} \frac{H_{2n}-H_n}{n}+\frac{1}{2}\sum_{n=1}^{\infty} \frac{y^{2n}}{n^2}$ where we divide both sides by $y$ and then integrate from $y=0$ to $y=x$, we have $\displaystyle -\int_0^x \frac{\log(1+y)\log(1-y)}{y}\textrm{d}y=\sum_{n=1}^{\infty} x^{2n} \frac{H_{2n}-H_n}{2n^2}+\frac{1}{4}\sum_{n=1}^{\infty} \frac{x^{2n}}{n^3}$. Now, if we multiply both sides of this last result by $\log(1+x)/x$ and then integrate from $x=0$ to $x=1$, using the fact that $\displaystyle \int_0^1 x^{2n-1}\log(1+x) \textrm{d}x=\frac{H_{2n}-H_n}{2n}$, we get

{A specific note: one can multiply both sides of the relation above by $\log(1-x)/x$ instead of $\log(1+x)/x$ and use the integral, $\int_0^1 x^{n-1}\log(1-x)\textrm{d}x=-H_n/n$, but later in the process one might like to use though the version $\int_0^1 x^{2n-1}\log(1+x) \textrm{d}x$ to nicely get the calculations.}

$$\underbrace{-\int_0^1 \frac{\log(1+x)}{x}\left(\int_0^x \frac{\log(1+y)\log(1-y)}{y}\textrm{d}y\right)\textrm{d}x}_{\displaystyle I}=\frac{5}{4}\sum_{n=1}^{\infty}\frac{H_n^2}{n^3}+\frac{7}{8}\sum_{n=1}^{\infty}\frac{H_n}{n^4}-\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{n}}{n^4}-\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n^2}{n^3}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^3}.$$

Integrating by parts, the integral $I$ may be written as $5/16\zeta(2)\zeta(3)-\underbrace{\int_0^1 \frac{\text{Li}_2(-x) \log (1+x) \log (1-x)}{x} \textrm{d}x}_{J}$, and then we may write that $$\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^3}=2\int_0^1 \frac{\text{Li}_2(-x) \log (1+x) \log (1-x)}{x} \textrm{d}x-\frac{5}{8}\zeta(2)\zeta(3)+\frac{5}{2}\sum_{n=1}^{\infty}\frac{H_n^2}{n^3}+\frac{7}{4}\sum_{n=1}^{\infty}\frac{H_n}{n^4}-2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{n}}{n^4}-2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n^2}{n^3}\tag1 .$$

Now, the last magical part comes from considering expressing the integral $J$ in a different way, and by using the Cauchy product, $\displaystyle \operatorname{Li}_2(-x)\log(1+x)=3\sum_{n=1}^{\infty}(-1)^n \frac{x^n}{n^3}-2 \sum_{n=1}^{\infty}(-1)^n x^n\frac{H_n}{n^2}-\sum_{n=1}^{\infty}(-1)^nx^n\frac{H_n^{(2)}}{n}$, we get that

$$\int_0^1 \frac{\operatorname{Li}_2(-x) \log (1+x) \log (1-x)}{x} \textrm{d}x= -\sum _{n=1}^{\infty }(-1)^{n-1}\frac{ H_n H_n^{(2)}}{n^2}+3\sum _{n=1}^{\infty }(-1)^{n-1}\frac{ H_n}{n^4}\\-2\sum _{n=1}^{\infty }(-1)^{n-1}\frac{H_n^2}{n^3}.\tag2$$

Combining $(1)$ and $(2)$ and collecting the values of the series from the book, (Almost) Impossible Integrals, Sums, and Series, we are done with the first series.

To get the value of the second series we might use the relation:

\begin{equation*} \sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n-1)^3}-\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n)^3} \end{equation*} \begin{equation*} =6 \log (2)-2 \log ^2(2)-\frac{1}{12}\log ^4(2)+\frac{3}{20} \log ^5(2)-\frac{3}{2} \zeta (2)-\frac{21}{8} \zeta (3)+\frac{173}{32} \zeta (4) \end{equation*} \begin{equation*} +\frac{55}{32} \zeta (5)-\frac{5 }{4}\zeta (2) \zeta (3)+\frac{3}{2} \log (2) \zeta (2)-\frac{7}{2}\log (2)\zeta (3)-4\log (2)\zeta (4)+\frac{1}{2} \log ^2(2) \zeta (2) \end{equation*} \begin{equation*} -\frac{5}{6} \log ^3(2)\zeta (2)+\frac{21}{8}\log ^2(2)\zeta (3)-2 \operatorname{Li}_4\left(\frac{1}{2}\right)+4 \log (2)\operatorname{Li}_4\left(\frac{1}{2}\right)+2 \operatorname{Li}_5\left(\frac{1}{2}\right), \end{equation*}

and this is obtained by using a very similar strategy to the one given in Section 6.59, pages $530$-$532$, from the book, (Almost) Impossible Integrals, Sums, and Series. The critical identity here is given in (6.289).

A detailed solution will appear soon in a new paper.

UPDATE (September $30$, $2019$)

A magical way to the series $\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}$ by Cornel Ioan Valean

By the Cauchy product, we have $\operatorname{Li}_2(x^2) \log(1-x^2)= 3\sum _{n=1}^{\infty } \frac{x^{2 n}}{n^3}-2\sum _{n=1}^{\infty } x^{2n}\frac{H_n}{n^2}-\sum _{n=1}^{\infty } x^{2n}\frac{H_n^{(2)}}{n}$, and if we multiply both sides by $\log(1-x)/x$, and integrate from $x=0$ to $x=1$, using that $\int_0^1 x^{n-1}\log(1-x)\textrm{d}x=-H_n/n$, and doing all the reductions, we arrive at

$$2\sum _{n=1}^{\infty } \frac{H_{2 n} H_n^{(2)}}{(2 n)^2}-12\sum _{n=1}^{\infty } \frac{H_n}{n^4}+12\sum _{n=1}^{\infty }(-1)^{n-1} \frac{H_n}{n^4}+\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^3}$$ $$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log \left(1-x^2\right) \log (1-x)}{x} \textrm{d}x$$ $$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log (1+x) \log (1-x)}{x} \textrm{d}x+2 \int_0^1 \frac{\text{Li}_2(-x) \log ^2(1-x)}{x} \textrm{d}x\\+2 \int_0^1 \frac{\text{Li}_2(x) \log ^2(1-x)}{x} \textrm{d}x$$ $$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log (1+x) \log (1-x)}{x} \textrm{d}x+2 \sum _{n=1}^{\infty } \frac{H_n^2}{n^3}-2 \sum _{n=1}^{\infty } \frac{(-1)^{n-1}H_n^2}{n^3}+2 \sum _{n=1}^{\infty } \frac{H_n^{(2)}}{n^3}\\-2 \sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_n^{(2)}}{n^3},$$ where the last integral is given here Two very advanced harmonic series of weight $5$, and all the last resulting harmonic series are given in the book (Almost) Impossible Integrals, Sums, and Series. The reduction to the last series has been achieved by using the identity, $\displaystyle \int_0^1 x^{n-1}\log^2(1-x)\textrm{d}x=\frac{H_n^2+H_n^{(2)}}{n}$. The series $\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^3}$ maybe found calculated in the paper On the calculation of two essential harmonicseries with a weight 5 structure, involving harmonic numbers of the type H_{2n} by Cornel Ioan Valean. Thus, we have


All the details will appear in a new paper.

UPDATE (October $30$, $2019$) The details with respect to the evaluation of the previous series may be found in the preprint The evaluation of a special harmonic series with a weight $5$ structure, involving harmonic numbers of the type $H_{2n}$

UPDATE (July $05$, $2020$) The two essential series stated above may be found now as an article here.

A very simple solution to one of the key logarithmic integrals stated above, $ \displaystyle \int_{0}^{1}x^{2n-1}\ln(1+x)\textrm{d}x=\frac{H_{2n}-H_n}{2n}$

Solution (by Cornel): \begin{equation*} \int_{0}^{1}x^{2n-1}\ln(1+x)\textrm{d}x= \int_0^1 \left(\frac{x^{2n}}{2n}-\frac{1}{2n}\right)' \log(1+x)\textrm{d}x=\frac{1}{2n}\int_0^1 \frac{1-x^{2n}}{1+x}\textrm{d}x \end{equation*} \begin{equation*} =\frac{1}{2n}\sum _{k=1}^{2 n} (-1)^{k-1} \int_0^1 x^{k-1} \textrm{d}x=\frac{1}{2n}\sum _{k=1}^{2 n}\frac{ (-1)^{k-1}}{k}=\frac{H_{2n}-H_n}{2n}. \end{equation*}

BONUS: A very simple solution to a logarithmic integral strongly related to the key integral above, $ \displaystyle \int_{0}^{1}x^{2n}\ln(1+x)\textrm{d}x=\frac{2\log(2)}{2n+1}-\frac{1}{(2n+1)^2}+\frac{H_n-H_{2n}}{2n+1}$

Solution (by Cornel): \begin{equation*} \int_{0}^1 x^{2n}\ln(1+x)\textrm{d}x= \int_{0}^1 \left(\frac{x^{2n+1}}{2n+1}+\frac{1}{2n+1}\right)' \ln(1+x)\textrm{d}x \end{equation*} \begin{equation*} =\frac{2\log(2)}{2n+1}-\frac{1}{2n+1}\int_0^1\frac{1+x^{2n+1}}{1+x}\textrm{d}x=\frac{2\log(2)}{2n+1}-\frac{1}{2n+1}\sum_{k=1}^{2n+1} (-1)^{k-1}\int_0^1x^{k-1}\textrm{d}x \end{equation*} \begin{equation*} =\frac{2\log(2)}{2n+1}-\frac{1}{2n+1}\sum_{k=1}^{2n+1} \frac{(-1)^{k-1}}{k}=\frac{2\log(2)}{2n+1}-\frac{1}{(2n+1)^2}+\frac{H_n-H_{2n}}{2n+1}. \end{equation*}

  • 1
    $\begingroup$ (+1) for sharing such awesome stuff. $\endgroup$ Sep 5 '19 at 17:14
  • $\begingroup$ @AliShather Thank you. $\endgroup$ Sep 6 '19 at 7:54

We have

$$\frac{\ln^2(1-y)}{1-y}=\sum_{n=1}^\infty y^n(H_n^2-H_n^{(2)})\tag{1}$$

integrate both sides of (1) from $y=0$ to $y=x$ to get


Now replace $x$ with $x^2$ in (2) then multiply both sides by $-\frac{\ln(1-x)}{x}$ and integrate from $x=0$ to $x=1$, also note that $\int_0^1 -x^{2n-1}\ln(1-x)\ dx=\frac{H_{2n}}{2n}$ we get

$$\frac13\underbrace{\int_0^1\frac{\ln^3(1-x^2)\ln(1-x)}{x}\ dx}_{\large I}=\sum_{n=1}^\infty\frac{H_{2n}}{2n^2}\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac{2}{n^2}\right)$$

Rearranging the terms to get


@nospoon mentioned in equation (3) of his solution that he found

$$\sum_{n=1}^{\infty} \frac{H_{n-1}^{(2)}\,H_{2n}}{n^2} =\frac{11}{4}\zeta(2)\,\zeta(3)-\frac{47}{16}\zeta(5)$$


$$\boxed{\sum _{n=1}^{\infty } \frac{H_{2n} H_{n}^{(2)}}{(2 n)^2}=\frac{101 }{64}\zeta (5)-\frac{5 }{16}\zeta (2) \zeta (3)}$$

Also Cornel elegantly calculated the second sum above

$$\boxed{\small{\sum _{n=1}^{\infty } \frac{H_{2 n}H_n }{(2 n)^3}=\frac{307}{128}\zeta(5)-\frac{1}{16}\zeta (2) \zeta (3)+\frac{1}{3}\ln ^3(2)\zeta (2) -\frac{7}{8} \ln ^2(2)\zeta (3)-\frac{1}{15} \ln ^5(2) -2 \ln (2) \operatorname{Li}_4\left(\frac{1}{2}\right) -2 \operatorname{Li}_5\left(\frac{1}{2}\right)}}$$

For the third sum: $$\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}=\frac12\sum_{n=1}^\infty\frac{H_{n}}{n^4}+\frac12\sum_{n=1}^\infty(-1)^n\frac{H_{n}}{n^4}$$

plugging the common results:



we get


For the remaining integral $I$, we use the magical identity


with $a=\ln(1-x)$ and $b=\ln(1+x)$ we can write

$$I=\int_0^1\frac{\ln^4(1-x)}{x}\ dx-\int_0^1\frac{\ln^4(1+x)}{x}\ dx+\frac12\underbrace{\int_0^1\frac{\ln^4(1-x^2)}{x}\ dx}_{x^2\mapsto x}\\-\underbrace{\frac12\int_0^1\frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}-\underbrace{\int_0^1\frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln(1+x)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}$$

$$I=\frac54\underbrace{\int_0^1\frac{\ln^4(1-x)}{x}\ dx}_{4!\zeta(5)}-\underbrace{\int_0^1\frac{\ln^4(1+x)}{x}\ dx}_{K}-\underbrace{\int_0^1\frac{\ln^4x}{1-x^2}\ dx}_{\frac{93}{4}\zeta(5)}+\underbrace{2\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx}_{J}$$


we have

\begin{align} K&=\int_0^1\frac{\ln^4(1+x)}{x}=\int_{1/2}^1\frac{\ln^4x}{x}\ dx+\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx\\ &=\frac15\ln^52+\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^4x\ dx\\ &=\frac15\ln^52+\sum_{n=1}^\infty\left(\frac{24}{n^5}-\frac{24}{n^52^n}-\frac{24\ln2}{n^42^n}-\frac{12\ln^22}{n^32^n}-\frac{4\ln^32}{n^22^n}-\frac{\ln^42}{n2^n}\right)\\ &=4\ln^32\zeta(2)-\frac{21}2\ln^22\zeta(3)+24\zeta(5)-\frac45\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right) \end{align}


$$J=2\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx=\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x}\ dx+\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1+x}\ dx$$

using the rule

$$\int_0^1\frac{\ln^ax\ln\left(\frac{1+x}{2}\right)}{1-x}\ dx=(-1)^aa!\sum_{n=1}^\infty\frac{(-1)^nH_n^{a+1}}{n}$$

allows us to write

\begin{align} J&=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}+\int_0^1\frac{\ln^3x\ln(1+x)}{1+x}\ dx-\ln2\int_0^1\frac{\ln^3x}{1+x}\ dx\\ &=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}-\sum_{n=1}^\infty(-1)^n H_n\int_0^1x^n\ln^3x\ dx-\ln2\left(-\frac{21}4\zeta(4)\right)\\ &=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}+6\sum_{n=1}^\infty\frac{(-1)^n H_n}{(n+1)^4}+\frac{21}{4}\ln2 \zeta(4)\\ &=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}-6\sum_{n=1}^\infty\frac{(-1)^n H_n}{n^4}-\frac{45}{8}\zeta(5)+\frac{21}{4}\ln2 \zeta(4) \end{align}



we get


Plugging the results of $K$ and $J$ in (4) we get


and finally by substituting the boxed results in (3) we get

$$\sum _{n=1}^{\infty } \frac{H_{2 n}H_n^2 }{(2 n)^2} =\frac{9 }{16}\zeta (2) \zeta (3)+\frac{421 }{64}\zeta (5)+\frac{2}{3} \ln ^3(2)\zeta (2) -\frac{7}{4} \ln ^2(2)\zeta (3)\\ -\frac{2}{15} \ln^5(2) -4 \ln2\operatorname{Li}_4\left(\frac{1}{2}\right) -4 \operatorname{Li}_5\left(\frac{1}{2}\right)$$


$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$ can be found here and $\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}$ can be found here.

  • $\begingroup$ (+1) Thanks. However, I specified clearly that I only need references. $\endgroup$ Sep 9 '19 at 11:26
  • $\begingroup$ Thank you and sorry I didnt read that part. $\endgroup$ Sep 9 '19 at 14:08

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