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How to prove that

\begin{align} \sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}=\frac13\ln^42-2\ln^22\zeta(2)+7\ln2\zeta(3)-\frac{121}{16}\zeta(4)+8\operatorname{Li}_4\left(\frac12\right) \end{align} where $H_n^{(m)}=1+\frac1{2^m}+\frac1{3^m}+...+\frac1{n^m}$ is the $n$th harmonic number of order $m$.

This problem was proposed by Cornel Valean.

Here is integral expression of the sum $\displaystyle -\int_0^1\frac{\ln x\operatorname{Li}_2(x^2)}{1-x^2}\ dx\quad $.

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  • $\begingroup$ I haven't tried it myself, but using the Cauchy product of two series should allow you to derive the closed form. $\endgroup$
    – uhhhhidk
    Jun 11, 2019 at 19:52
  • $\begingroup$ @uhhhhidk sounds a good idea but I think that will make it more complicated. would like to see your approach. $\endgroup$ Jun 11, 2019 at 19:56
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    $\begingroup$ I tried that, and after considerable work to simplify, I got back the original expression. $\endgroup$ Jun 11, 2019 at 20:22
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    $\begingroup$ @Zacky coz harmonic sums are very related to logarithmic/Polylogarithmic integrals. And I mentioned in the body that I tried to solve it using integration but got complicated. I am going to edit the post and add the integral representation. $\endgroup$ Jun 11, 2019 at 20:54
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    $\begingroup$ @Zachy no problem. this problem was proposed by Cornel in his FB page and i solved it relying on a value of integral evaluated by a friend and the value of $\sum_{n=1}^\infty \frac{(-1)^nH_n}{(2n+1)^3$ which was proposed by Cornel too and i managed to calculate. see the link here facebook.com/… $\endgroup$ Jun 11, 2019 at 21:21

2 Answers 2

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A solution using Abel's summation as suggested by Cornel.

Let $\ \displaystyle S\ $ denote $\ \displaystyle \sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}\ $ and by using Abel's summation:

$\displaystyle\sum_{k=1}^n a_k b_k=A_nb_{n+1}+\sum_{k=1}^{n}A_k\left(b_k-b_{k+1}\right)\ $ where $\ \displaystyle A_n=\sum_{i=1}^n a_i\ $

and letting let $\ \displaystyle a_k=\frac{1}{(2k+1)^2}\ $ , $\ \displaystyle b_k=H_k^{(2)}$, we get

\begin{align} \sum_{k=1}^n\frac{H_k^{(2)}}{(2k+1)^2}&=\sum_{i=1}^n\frac{H_{n+1}^{(2)}}{(2i+1)^2}-\sum_{k=1}^n\frac{1}{(k+1)^2}\left(\sum_{i=1}^k\frac{1}{(2i+1)^2}\right)\\ &=\sum_{i=1}^n\frac{H_{n+1}^{(2)}}{(2i+1)^2}-\sum_{k=1}^n\frac{1}{(k+1)^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}+\frac{1}{(2k+1)^2}-1\right) \end{align} Letting $n$ approach $\infty$, we get \begin{align} S&=\zeta(2)\sum_{i=1}^\infty\frac{1}{(2i+1)^2}-\sum_{k=1}^\infty\frac{1}{(k+1)^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)\\ &\quad-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}+\sum_{k=1}^\infty\frac1{(k+1)^2}\\ &=\zeta(2)\left(\frac34\zeta(2)-1\right)-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}-\frac{1}{(2k-1)^2}\right)\\ &\quad-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}+\zeta(2)-1\\ &=\frac{15}8\zeta(4)-1-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)+\sum_{k=1}^\infty\frac{1}{k^2(2k-1)^2}-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}\\ &=\frac{15}8\zeta(4)-1-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)+1\\ &\quad+\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}\\ &=\frac{15}8\zeta(4)-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)\\ &=\frac{15}8\zeta(4)-4\sum_{k=1}^\infty\frac{H_{2k}^{(2)}}{(2k)^2}+\frac14\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}\\ &=\frac{15}8\zeta(4)-4\left(\frac12\sum_{k=1}^\infty\frac{H_{k}^{(2)}}{k^2}+\frac12\sum_{k=1}^\infty\frac{(-1)^kH_k^{(2)}}{k^2}\right)+\frac14\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}\\ &=\frac{15}8\zeta(4)-\frac74\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}-2\sum_{k=1}^\infty\frac{(-1)^kH_k^{(2)}}{k^2} \end{align} By plugging $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^nH_n^{(2)}}{n^2}=-4\operatorname{Li}_4\left(\frac12\right)+\frac{51}{16}\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac16\ln^42\ $

( proved here ) and $\ \displaystyle\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}=\frac74\zeta(4)\ $, we get the closed form of $\ S$

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Different approach:

\begin{align} S&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{(2n+1)^2}\\ &=\sum_{n=1}^\infty H_n^{(2)}\int_0^1-x^{2n}\ln x\ dx\\ &=-\int_0^1\ln x\sum_{n=1}^\infty(x^2)^nH_n^{(2)}\\ &=-\int_0^1\frac{\ln x\operatorname{Li}_2(x^2)}{1-x^2}\ dx,\quad \operatorname{Li}_2(x^2)=2\operatorname{Li}_2(x)+2\operatorname{Li}_2(-x)\\ &=-2\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1-x^2}\ dx-2\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1-x^2}\ dx\\ &=-\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1-x}-\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1+x}-\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1-x}-\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1+x}\ dx\\ &=-I_1-I_2-I_3-I_4 \end{align}


\begin{align} I_1&=\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1-x}\ dx\\ &=\sum_{n=1}^\infty H_n^{(2)}\int_0^1 x^n\ln x\ dx\\ &=-\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2}\\ &=-\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^2}+\zeta(4) \end{align}


\begin{align} I_2&=\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1+x}\ dx\\ &=-\sum_{n=1}^\infty (-1)^n\int_0^1\ x^{n-1}\ln x\operatorname{Li}_2(x)\ dx\\ &=-\sum_{n=1}^\infty (-1)^n\left(\frac{H_n^{(2)}}{n^2}+\frac{2H_n}{n^3}-\frac{2\zeta(2)}{n^2}\right) \end{align}


\begin{align} I_3&=\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(H_n^{(2)}-\zeta(2)\right) \end{align}


\begin{align} I_4&=\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1+x}\ dx\\ &=-\sum_{n=1}^\infty (-1)^nH_n^{(2)}\int_0^1x^n\ln x\ dx\\ &=-\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{(n+1)^2}\\ &=\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{n^2}+\frac78\zeta(4) \end{align}


Combine the four integrals we get

$$S=\frac98\zeta(4)+2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^3}-\sum_{n=1}^\infty(-1)^n\frac{H_n^{(2)}}{n^2}$$

Plugging the two sums we get

$$S=\frac13\ln^42-2\ln^22\zeta(2)+7\ln2\zeta(3)-\frac{121}{16}\zeta(4)+8\operatorname{Li}_4\left(\frac12\right) $$

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  • $\begingroup$ Maybe you can combine answers? $\endgroup$ Oct 19, 2021 at 21:45

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