# $\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)$

This problem was proposed by Cornel and he showed that

$$S=\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)=\frac13\ln^42-\frac12\ln^22\zeta(2)+\frac72\ln2\zeta(3)-\frac{21}4\zeta(4)+8\operatorname{Li}_4\left(\frac12\right)$$

here is my approach.

we know that $$\quad\zeta(2)-H_{2n}^{(2)}=\psi^{(1)}(2n+1)=-\displaystyle\int_0^1\frac{x^{2n}\ln x}{1-x}\ dx$$

then $$S=-\int_0^1\frac{\ln x}{1-x}\sum_{n=1}^\infty\frac{x^{2n}H_n}{n}\ dx=-\int_0^1\frac{\ln x}{1-x}\left(\frac12\ln^2(1-x^2)+\operatorname{Li}_2(x^2)\right)\ dx$$ and by applying IBP , we get $$S=-2\int_0^1\frac{\operatorname{Li}_2(1-x)\ln(1-x^2)}{x(1-x^2)}\ dx$$ I applied the dilogarithmic identity $$\quad\operatorname{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)$$

but it was not that helpful. any idea?

in the body, I reached \begin{align} S&=-2\int_0^1\frac{\operatorname{Li}_2(1-x)\ln(1-x^2)}{x(1-x^2)}\ dx\\ &=-2\int_0^1\frac{\operatorname{Li}_2(1-x)\left[\ln(1-x)+\ln(1+x)\right]}{x(1-x)(1+x)}\ dx,\qquad 1-x=y\\ &=-2\int_0^1\frac{\operatorname{Li}_2(x)\left[\ln x+\ln(2-x)\right]}{x(1-x)(2-x)}\ dx\\ &=-2\int_0^1\frac{\operatorname{Li}_2(x)\ln x}{x(1-x)(2-x)}\ dx-2\int_0^1\frac{\operatorname{Li}_2(x)\ln(2-x)}{x(1-x)(2-x)}\ dx\\ &=-2\left(I_1+I_2\right) \end{align} using the partial fraction decomposition $$\quad\displaystyle\frac{1}{x(1-x)(2-x)}=\frac1{1-x}+\frac12\left(\frac1{x}-\frac1{2-x}\right)$$, we get \begin{align} I_1&=\int_0^1\frac{\operatorname{Li}_2(x)\ln x}{1-x}\ dx+\frac12\int_0^1\frac{\operatorname{Li}_2(x)\ln x}{x}\ dx-\frac12\int_0^1\frac{\operatorname{Li}_2(x)\ln x}{2-x}\ dx\\ &=\sum_{n=1}^\infty\frac1{n^2}\int_0^1\frac{x^n\ln x}{1-x}\ dx+\frac12\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^{n-1}\ln x\ dx-\frac12\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\ln x\operatorname{Li}_2(x)\ dx\\ &=-\sum_{n=1}^\infty\frac{\zeta(2)-H_n^{(2)}}{n^2}-\frac12\sum_{n=1}^\infty\frac1{n^4}-\frac12\sum_{n=1}^\infty\frac1{2^n}\left(\frac{2H_n}{n^3}+\frac{H_n^{(2)}}{n^2}-\frac{2\zeta(2)}{n^2}\right)\\ &=-\zeta^2(2)+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^2}-\frac12\zeta(4)-\sum_{n=1}^\infty\frac{H_n}{n^32^n}-\frac12\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}+\zeta(2)\operatorname{Li}_2\left(\frac12\right) \end{align} plugging the value of the second and third sum proved in here and here respectively, along with the value of the first sum $$\frac74\zeta(4)$$ and $$\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$$, we get $$\color{red}{I_1=-\frac1{16}\ln^42-\frac38\ln^22\zeta(2)-\frac{5}{32}\zeta(4)-\frac32\operatorname{Li}_4\left(\frac12\right)}$$ using the partial fraction decomposition for the second integral too, we get, \begin{align} I_2&=\int_0^1\frac{\operatorname{Li}_2(x)\ln(2-x)}{1-x}\ dx+\frac12\int_0^1\frac{\operatorname{Li}_2(x)\ln(2-x)}{x}\ dx-\frac12\int_0^1\frac{\operatorname{Li}_2(x)\ln(2-x)}{2-x}\ dx\\ &=A+\frac12B-\frac12C \end{align} To evaluate $$A\$$, let $$1-x=y$$ then apply IBP, we get \begin{align} A&=\int_0^1\frac{\operatorname{Li}_2(x)\ln(2-x)}{1-x}\ dx=\int_0^1\frac{\operatorname{Li}_2(1-x)\ln(1+x)}{x}\ dx\\ &=\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1-x}\ dx=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1\frac{x^n\ln x}{1-x}\ dx\\ &=-\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\left(\zeta(2)-H_n^{(2)}\right)\\ &=-\zeta(2)\operatorname{Li}_2(-1)+\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^2} \end{align} substituting the closed form of the sum calculated here along with $$\operatorname{Li}_2(-1)=-\frac12\zeta(2)$$, we get

$$\color{blue}{A=-\frac16\ln^42+\ln^22\zeta(2)-\frac72\ln2\zeta(3)+\frac{71}{16}\zeta(4)-4\operatorname{Li}_4\left(\frac12\right)}$$ \begin{align} B&=\int_0^1\frac{\operatorname{Li}_2(x)\ln(2-x)}{x}\ dx=\int_0^1\frac{\operatorname{Li}_2(x)\left[\ln2+\ln(1-x/2)\right]}{x}\ dx\\ &=\ln2\operatorname{Li}_3(1)-\sum_{n=1}^\infty\frac{1}{n2^n}\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\\ &=\ln2\zeta(3)-\sum_{n=1}^\infty\frac{1}{n2^n}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)\\ &=\ln2\zeta(3)-\zeta(2)\operatorname{Li}_2\left(\frac12\right)+\sum_{n=1}^\infty\frac{H_n}{n^32^n} \end{align} plugging the well known result of the sum and $$\operatorname{Li}_2\left(\frac12\right)$$, we get $$\color{blue}{B=\frac1{24}\ln^42+\frac12\ln^22\zeta(2)+\frac78\ln2\zeta(3)-\frac{9}{8}\zeta(4)+\operatorname{Li}_4\left(\frac12\right)}$$ \begin{align} C&=\int_0^1\frac{\operatorname{Li}_2(x)\ln(2-x)}{2-x}\ dx=\int_0^1\frac{\operatorname{Li}_2(x)\left[\ln2+\ln(1-x/2)\right]}{2-x}\ dx\\ &=\ln2\int_0^1\frac{\operatorname{Li}_2(x)}{2-x}\ dx+\frac12\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x/2)}{1-x/2}\ dx\\ &=\ln2\sum_{n=1}^\infty\frac{1}{2^n}\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx-\sum_{n=1}^\infty\left(\frac{H_n}{2^n}-\frac{1}{2^n}\right)\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\\ &=\ln2\sum_{n=1}^\infty\frac{1}{2^n}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)-\sum_{n=1}^\infty\left(\frac{H_n}{2^n}-\frac{1}{n2^n}\right)\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)\\ &=\sum_{n=1}^\infty\frac{H_n^2}{n^22^n}-\sum_{n=1}^\infty\frac{H_n}{n^32^n}-\ln2\sum_{n=1}^\infty\frac{H_n}{n^22^n}-\zeta(2)\sum_{n=1}^\infty\frac{H_n}{n2^n}+\ln^22\zeta(2)+\zeta(2)\operatorname{Li}_2\left(\frac12\right) \end{align} substituting the value of the first sum ( can be found here ) and the values of the other well known sums, we get $$\color{blue}{C=-\frac1{12}\ln^42+\frac54\ln^22\zeta(2)-\frac{21}8\ln2\zeta(3)+\frac{35}{16}\zeta(4)-2\operatorname{Li}_4\left(\frac12\right)}$$ by grouping $$A$$, $$B$$ and $$C$$, we get $$\color{red}{I_2=-\frac5{48}\ln^42+\frac58\ln^22\zeta(2)-\frac74\ln2\zeta(3)+\frac{89}{32}\zeta(4)-\frac52\operatorname{Li}_4\left(\frac12\right)}$$ Finally \begin{align} S&=-2\left(I_1+I_2\right)\\ &=\frac13\ln^42-\frac12\ln^22\zeta(2)+\frac72\ln2\zeta(3)-\frac{21}4\zeta(4)+8\operatorname{Li}_4\left(\frac12\right) \end{align}

• (+1) Interesting solution. Oct 29, 2020 at 21:12

A solution by Cornel Ioan Valean

If we set $$a_n=H_n/n$$ and $$b_n=\zeta(2)-H_{2n}^{(2)}$$, and then apply Abel's summation, we arrive at $$\sum_{n=1}^{\infty}\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)$$ $$=\frac{1}{2}\underbrace{\sum_{n=1}^{\infty}\frac{H_n^2-H_n^{(2)}}{(2n+1)^2}}_{\displaystyle A}+\underbrace{\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{(2n+1)^2}}_{\displaystyle B}+\frac{1}{8}\underbrace{\sum_{n=1}^{\infty}\frac{H_n^2+H_n^{(2)}}{n^2}}_{\displaystyle C}-\frac{1}{4}\underbrace{\sum_{n=1}^{\infty}\frac{H_n}{n^3}}_{\displaystyle 5/16\zeta(3)}.\tag1$$

For the sum $$A$$, we consider a result from the book (Almost) Impossible Integrals, Sums, and Series, page $$291$$, $$\begin{equation*} A=\sum_{k=1}^{\infty} \frac{H_k^2-H_k^{(2)}}{(k+1)(k+n+1)} \end{equation*}$$ $$\begin{equation*} =\frac{(\psi(n+1)+\gamma)^3+3(\psi(n+1)+\gamma)(\zeta(2)-\psi^{(1)}(n+1))+2(\zeta(3)+1/2\psi^{(2)}(n+1))}{3n}, \end{equation*}$$ where if we multiply both sides by $$n$$, differentiate once with respect to $$n$$, and set $$n=-1/2$$, we arrive at $$\begin{equation*} \sum_{n=1}^{\infty} \frac{H_n^2-H_n^{(2)}}{(2n+1)^2}=\frac{15}{4}\zeta (4)-7 \log(2) \zeta(3)+3\log^2(2)\zeta(2). \end{equation*}$$

For the sum $$B$$, we apply Abel's summation with $$a_n=1/(2n+1)^2$$ and $$H_n^{(2)}$$ that gives $$\begin{equation*} B=\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{(2n+1)^2}=\frac{15}{8}\zeta(4)-\frac{7}{4}\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^2}+2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n^{(2)}}{n^2} \end{equation*}$$ $$\begin{equation*} =\frac{1}{3}\log^4(2)-2\log^2(2)\zeta(2)+7\log(2)\zeta(3)-\frac{121}{16}\zeta(4)+8 \operatorname{Li}_4\left(\frac{1}{2}\right), \end{equation*}$$ where $$\begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2}=\frac{1}{6}\log^4(2)-\log^2(2)\zeta(2)+\frac{7}{2}\log(2)\zeta(3)-\frac{51}{16}\zeta(4)+4\operatorname{Li}_4\left(\frac{1}{2}\right), \end{equation*}$$ may be found calculated in (Almost) Impossible Integrals, Sums, and Series, pages $$505$$-$$506$$.

Finally, for the sum $$C$$ we may elegantly exploit the fact that $$\displaystyle\int_0^1 x^{n-1}\log^2(1-x)\textrm{d}x =\frac{H_n^2+H_n^{(2)}}{n}$$, you may also find nicely calculated in (Almost) Impossible Integrals, Sums, and Series, page $$60$$. In general, such integrals may also be viewed in terms of Beta function and hence a solution is pretty easily extracted. Then, $$C=\sum_{n=1}^{\infty}\frac{H_n^2+H_n^{(2)}}{n^2}=-\int_0^1 \frac{\log^3(1-x)}{x}\textrm{d}x=-\int_0^1 \frac{\log^3(x)}{1-x}\textrm{d}x=6\zeta(4).$$

Collecting all sums in $$A$$, $$B$$, $$C$$ and plugging them in $$(1)$$, we arrive at the desired result $$\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)=\frac{1}{3}\log^4(2)-\frac{1}{2}\log^2(2)\zeta(2)+\frac{7}{2}\log(2)\zeta(3)-\frac{21}{4}\zeta(4)+8\operatorname{Li}_4\left(\frac{1}{2}\right).$$

COOL BONUS

Combining the sums $$A$$ and $$B$$ we obtain that $$\sum_{n=1}^{\infty} \frac{H_n^2}{(2n+1)^2}=\frac{1}{3}\log^4(2)+\log^2(2)\zeta(2)-\frac{61}{16}\zeta(4)+8 \operatorname{Li}_4\left(\frac{1}{2}\right).$$