5
$\begingroup$

In a post found here it is mentioned that a closed form for the so-called younger brother (younger in the sense the power in the denominator is only squared, rather than cubed as in the linked question) skew-harmonic sum $$S = \sum_{n = 1}^\infty \frac{H_n \overline{H}_n}{n^2}$$ can be found, though none is given. Here $H_n = \sum_{k = 1}^n \frac{1}{k}$ is the $n$th harmonic number while $\overline{H}_n = \sum_{k = 1}^n \frac{(-1)^{k + 1}}{k}$ is the $n$th skew-harmonic number.

I seek the closed-form expression for the sum $S$.

My thoughts on a possible alternative approach to that suggested in the link is as follows. Since $$\ln 2 - \overline{H}_n = (-1)^n \int_0^1 \frac{x^n}{1 + x} \, dx,$$ then $$H_n \overline{H}_n = \ln 2 H_n -(-1)^n H_n \int_0^1 \frac{x^n}{1 + x} \, dx.$$ Thus \begin{align} S &= \ln 2 \sum_{n = 1}^\infty \frac{H_n}{n^2} - \int_0^1 \frac{1}{1 + x} \sum_{n = 1}^\infty \frac{(-1)^n H_n x^n}{n^2} \, dx\\ &= 2 \ln 2 \zeta (3) - \int_0^1 \frac{1}{1 + x} \sum_{n = 1}^\infty \frac{H_n (-x)^n}{n^2} \, dx, \end{align} since $\sum_{n = 1}^\infty \frac{H_n}{n^2} = 2 \zeta (3)$.

I then thought of perhaps using the following known generating function of $$\sum_{n = 1}^\infty \frac{H_n}{n^2} x^n = \operatorname{Li}_3 (x) - \operatorname{Li}_3 (1-x) + \ln (1 - x) \operatorname{Li}_2 (1 - x) + \frac{1}{2} \ln x \ln^2 (1 - x) + \zeta (3),$$ but this leads to complex valued logs and polylogs which I would rather avoid.


Continuing

Continuing on using the generating function, we see that \begin{align} S &= 2 \ln 2 \zeta (3) - \int_0^1 \frac{\operatorname{Li}_3 (-x)}{1 + x} \, dx + \int_0^1 \frac{\operatorname{Li}_3 (1 + x)}{1 + x} \, dx\\ & \qquad - \int_0^1 \frac{\ln (1 + x) \operatorname{Li}_2 (1 + x)}{1 + x} \, dx - \frac{1}{2} \int_0^1 \frac{\ln (-x) \ln^2 (1 + x)}{1 + x} \, dx - \zeta (3) \int_0^1 \frac{dx}{1 + x}. \end{align} Surprisingly, indefinite integrals for all integrals appearing above can be readily found. Here: \begin{align} \int_0^1 \frac{\operatorname{Li}_3(-x)}{1 + x} \, dx &= \frac{1}{2} \operatorname{Li}^2_2 (-x) + \operatorname{Li}_3 (-x) \ln (1 + x) \Big{|}_0^1 = \frac{5}{16} \zeta (4) - \frac{3}{4} \ln 2 \zeta (3)\\[2ex] \int_0^1 \frac{\operatorname{Li}_3 (1 + x)}{1 + x} \, dx &= \operatorname{Li}_4 (1 + x) \Big{|}_0^1 = \operatorname{Li}_4 (2) - \zeta (4)\\ \int_0^1 \frac{\ln (1 + x) \operatorname{Li}_2 (1 + x)}{1 + x} \, dx &= \operatorname{Li}_3 (1 + x) \ln (1 + x) - \operatorname{Li}_4 (1 + x) \Big{|}_0^1\\ &= \operatorname{Li}_3 (2) \ln 2 - \operatorname{Li}_4 (2) + \zeta (4)\\[2ex] \int_0^1 \frac{\ln (-x) \ln^2 (1 + x)}{1 + x} \, dx &= -2 \operatorname{Li}_2 (1 + x) - \operatorname{Li}_2 (1 + x) \ln^2 (1 + x)\\ & \qquad + 2 \operatorname{Li}_3 (1 + x) \ln (1 + x) \Big{|}_0^1\\ &= -2 \operatorname{Li}_4 (2) - \operatorname{Li}_2 (2) \ln^2 2 + 2 \operatorname{Li}_3 (2) \ln 2 + 2 \zeta (4)\\ \int_0^1 \frac{dx}{1 + x} &= \ln 2 \end{align} Thus $$S = \frac{7}{4} \ln 2 \zeta (3) - \frac{53}{16} \zeta (4) + 3 \operatorname{Li}_4 (2) - 2 \operatorname{Li}_3 (2) \ln 2 + \frac{1}{2} \operatorname{Li}_2 (2) \ln^2 2.$$ Now finding values for $\operatorname{Li}_n (2)$ when $n = 2, 3$, and $4$. In each case the principal value is found.

  1. $n = 2$ case

Using $$\operatorname{Li}_2 (z) + \operatorname{Li}_2 (1 - z) = \zeta (2) - \ln z \ln (1 - z),$$ setting $z = 2$ gives $$\operatorname{Li}_2 (2) = \frac{3}{2} \zeta (2) - i\pi \ln 2.$$

  1. $n = 3$ case

Using $$\operatorname{Li}_3 (z) = \operatorname{Li}_3 \left (\frac{1}{z} \right ) - \frac{1}{6} \ln^3 (-z) - \zeta (2) \ln (-z),$$ setting $z = 2$ gives $$\operatorname{Li}_3 (2) = \frac{21}{24} \zeta (3) + \frac{3}{2} \zeta (2) \ln 2 - \frac{i \pi}{2} \ln^2 2.$$

  1. $n = 4$ case

Finally, from the result given here one has $$\operatorname{Li}_4(2) = 2 \zeta (4) - \operatorname{Li}_4 \left (\frac{1}{2} \right ) - \frac{i \pi}{6} \ln^3 2 + \zeta (2) \ln^2 2 - \frac{1}{24} \ln^4 2.$$

Plugging in all the pieces, we finally arrive at $$\sum_{n = 1}^\infty \frac{H_n \overline{H}_n}{n^2} = \frac{43}{16} \zeta (4) - 3 \operatorname{Li}_4 \left (\frac{1}{2} \right ) - \frac{1}{8} \ln^4 2 + \frac{3}{4} \zeta (2) \ln^2 2.$$ Magical!!

$\endgroup$
  • 1
    $\begingroup$ Just follow Cornels second solution in that link you provided. $\endgroup$ – Ali Shather Jan 26 at 8:40
  • 1
    $\begingroup$ @user97357329 I continued my generating function approach and, somewhat unbelievably, managed to arrive at the desired closed form. Truly amazing! $\endgroup$ – omegadot Jan 26 at 22:21
  • 1
    $\begingroup$ @AliShather see the magical way above. :-) $\endgroup$ – user97357329 Jan 26 at 22:29
  • 1
    $\begingroup$ @user97357329 elegant .. $\endgroup$ – Ali Shather Jan 27 at 0:34
  • 1
    $\begingroup$ @Ali Shather Thanks for picking up this simplification I missed. $\endgroup$ – omegadot Jan 28 at 20:10
4
$\begingroup$

Using Cornel's strategy here https://math.stackexchange.com/q/3459740, we have

$$\sum_{n=1}^{\infty} \frac{H_n \overline{H}_n}{n^2}=1+\sum_{n=1}^{\infty} \frac{H_{2n} \overline{H}_{2n}}{(2n)^2}+\sum_{n=1}^{\infty} \frac{H_{2n+1} \overline{H}_{2n+1}}{(2n+1)^2}$$ $$=1+\sum_{n=1}^{\infty}\frac{H_{2 n}^2}{(2n)^2}+\sum_{n=1}^{\infty}\frac{H_{2 n+1}^2}{(2 n+1)^2}-\sum_{n=1}^{\infty}\frac{H_n}{(2 n+1)^3}-\sum_{n=1}^{\infty}\frac{H_n H_{2 n}}{(2 n)^2}-\sum_{n=1}^{\infty}\frac{H_n H_{2 n}}{(2 n+1)^2}$$ $$=\frac{43}{16}\zeta(4)+\frac{3}{4}\log^2(2)\zeta(2)-\frac{1}{8}\log^4(2)-3\operatorname{Li}_4\left(\frac{1}{2}\right),$$

where the hardest sums (the last two ones) use the strategy in this paper On the calculation of two essential harmonic series with a weight 5 structure, involving harmonic numbers of the type $H_{2n}$ by C.I. Valean.

A first note: The hardest sums in closed-form are $$i) \ \sum_{n=1}^{\infty}\frac{H_n H_{2 n}}{(2 n)^2}=\frac{13}{32}\zeta(4)+\frac{7}{8}\log(2)\zeta(3)-\frac{1}{4}\log^2(2)\zeta(2)+\frac{1}{24}\log^4(2)+\operatorname{Li}_4\left(\frac{1}{2}\right);$$ $$ii) \ \sum_{n=1}^{\infty}\frac{H_n H_{2 n}}{(2 n+1)^2}=\frac{1}{12}\log^4(2)-\frac{1}{2}\log^2(2)\zeta(2)+\frac{7}{8}\log(2)\zeta(3)-\frac{1}{4}\zeta(4)+2\operatorname{Li}_4\left(\frac{1}{2}\right),$$ and are immediately extracted by following the strategy in the paper mentioned above. These results I took directly form Cornel's page here and here, to avoid boring calculations.

A second note: From the strategy in this answer here one may obtain immediately that $$\sum_{n=1}^{\infty}\frac{H_n}{(2 n+1)^3}=\frac{45}{32}\zeta(4)-\frac{7}{4}\log(2)\zeta(3).$$

Further, the first two sums from the second line may be immediately reduced to sums that are pretty known and they may also be found in the book, (Almost) Impossible Integrals, Sums, and Series.

A third note: The weight $5$ version of the present series may be found and calculated in the paper Two advanced harmonic series of weight 5 involving skew-harmonic numbers by C.I. Valean. As seen in the paper, the present series can also be approached in a different way.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Following the same approach here

$$S=\sum_{n=1}^\infty\frac{H_n\overline{H}_n}{n^2}=1+\sum_{n=2}^\infty\frac{H_n\overline{H}_n}{n^2}$$

Now use $$\sum_{n=2}^\infty f(n)=\sum_{n=1}^\infty f(2n)+\sum_{n=1}^\infty f(2n+1)$$

$$\Longrightarrow S=1+\frac14\sum_{n=1}^\infty\frac{H_{2n}\overline{H}_{2n}}{n^2}+\sum_{n=1}^\infty\frac{H_{2n+1}\overline{H}_{2n+1}}{(2n+1)^2}$$

$$=1+\frac14S_1+S_2$$

For $S_1$, use $\overline{H}_{2n}=H_{2n}-H_n$

$$S_1=\sum_{n=1}^\infty\frac{H_{2n}^2}{n^2}-\sum_{n=1}^\infty\frac{H_{2n}H_n}{n^2}$$

$$=2\sum_{n=1}^\infty\frac{H_{n}^2}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^nH_{n}^2}{n^2}-\sum_{n=1}^\infty\frac{H_{2n}H_n}{n^2}$$

For $S_2$, use $\overline{H}_{2n+1}=H_{2n+1}-H_n$

\begin{align} S_2&=\sum_{n=1}^\infty\frac{H_{2n+1}^2}{(2n+1)^2}-\sum_{n=1}^\infty\frac{H_{2n+1}H_n}{(2n+1)^2}\\ &=\sum_{n=0}^\infty\frac{H_{2n+1}^2}{(2n+1)^2}-1-\sum_{n=1}^\infty\frac{H_{2n}H_n}{(2n+1)^2}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^3}\\ &=\frac12\sum_{n=0}^\infty\frac{H_{n+1}^2}{(n+1)^2}+\frac12\sum_{n=0}^\infty\frac{(-1)^nH_{n+1}^2}{(n+1)^2}-1-\sum_{n=1}^\infty\frac{H_{2n}H_n}{(2n+1)^2}+\sum_{n=1}^\infty\frac{H_n}{(2n+1)^3}\\ &=\frac12\sum_{n=1}^\infty\frac{H_{n}^2}{n^2}-\frac12\sum_{n=1}^\infty\frac{(-1)^nH_{n}^2}{n^2}-1-\sum_{n=1}^\infty\frac{H_{2n}H_n}{(2n+1)^2}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^3}\\ \end{align}

and all these sums are known and the hardest ones are

$\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}=4\operatorname{Li_4}\left( \frac12\right)+\frac{13}{8}\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\ln^42$

and $\sum_{n=1}^\infty\frac{H_{2n}H_n}{(2n+1)^2}$ which I think was calculated by Cornel.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Different approach

From this paper in page $95$ Eq $(5)$ we have $$\sum_{n=1}^\infty \overline{H}_n\frac{x^n}{n}=\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2(-x)-\ln2\ln(1-x)-\operatorname{Li}_2\left(\frac12\right)$$

multiply both sides by $-\frac{\ln(1-x)}{x}$ then $\int_0^1$ and use the fact that $-\int_0^1x^{n-1}\ln(1-x)\ dx=\frac{H_n}{n}$ we get

$$\sum_{n=1}^\infty \frac{\overline{H}_nH_n}{n^2}=\underbrace{-\int_0^1\frac{\operatorname{Li}_2\left(\frac{1-x}{2}\right)\ln(1-x)}{x}\ dx}_{\large \mathcal{I}}+\underbrace{\int_0^1\frac{\operatorname{Li}_2(-x)\ln(1-x)}{x}\ dx}_{\large \mathcal{J}}$$ $$+\ln2\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ dx}_{2\zeta(3)}+\operatorname{Li}_2\left(\frac12\right)\underbrace{\int_0^1\frac{\ln(1-x)}{x}\ dx}_{-\zeta(2)}\tag1$$


Evaluation of $\mathcal{J}$

$$\int_0^1\frac{\operatorname{Li}_2(-x)\ln(1-x)}{x}\ dx=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{n-1}\ln(1-x)\ dx=-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$$ the last sum is calculated here

$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$$

$$\mathcal{J}=-2\operatorname{Li_4}\left(\frac12\right)+\frac{11}4\zeta(4)-\frac74\ln2\zeta(3)+\frac12\ln^22\zeta(2)-\frac{1}{12}\ln^42$$


Evaluation of $\mathcal{I}$

by Integration by parts twice we have

$$\mathcal{I}=\underbrace{-\int_0^1\frac{\ln(1-x)\operatorname{Li}_2(x)}{1+x}\ dx}_{\large \mathcal{I}_1}+\underbrace{\int_0^1\frac{\ln^2(1-x)\ln(1+x)}{x}\ dx}_{\large \mathcal{I}_2}-\ln2\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ dx}_{2\zeta(3)}$$

$\mathcal{I}_1$ was discussed here, using Landens identity $-\operatorname{Li}_2(x)=\operatorname{Li}_2\left(\frac{x}{x-1}\right)+\frac12\ln^2(1-x)$

$$\mathcal{I}_1=\int_0^1\frac{\ln(1-x)\operatorname{Li}_2\left(\frac{x}{x-1}\right)}{1+x}\ dx+\frac12\int_0^1\frac{\ln^3(1-x)}{1+x}\ dx$$

where the first integral is calculated here

$$\int_0^1 \frac{ \ln(1-x)\operatorname{Li}_2\left(\frac{x}{x-1}\right)}{1+x} \textrm{d}x=\frac{29}{16} \zeta (4)+\frac{1}{4}\ln ^22 \zeta (2) -\frac{1}{8} \ln ^42$$

and

$$\int_0^1\frac{\ln^3(1-x)}{1+x}\ dx=\int_0^1\frac{\ln^3y}{2-y}\ dy=\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\ln^3y \ dy=-6\sum_{n=1}^\infty\frac{1}{2^nn^4}=-6\operatorname{Li}_4\left(\frac12\right)$$

$$\Longrightarrow \mathcal{I}_1=\boxed{-3\operatorname{Li}_4\left(\frac12\right)+\frac{29}{16} \zeta (4)+\frac{1}{4}\ln ^22 \zeta (2) -\frac{1}{8} \ln ^42}$$


For $\mathcal{I}_2$, use $a^2b=\frac16(a+b)^3-\frac16(a-b)^3-\frac13b^3$ where $a=\ln(1-x)$ and $b=\ln(1+x)$

$$\Longrightarrow \mathcal{I}_2=\frac16\underbrace{\int_0^1\frac{\ln^3(1-x^2)}{x}\ dx}_{1-x^2=y}-\frac16\underbrace{\int_0^1\frac{\ln^3\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}=y}-\frac13\int_0^1\frac{\ln^3(1+x)}{x}\ dx$$

$$=\frac1{12}\underbrace{\int_0^1\frac{\ln^3y}{1-y}\ dy}_{-6\zeta(4)}-\frac13\underbrace{\int_0^1\frac{\ln^3y}{1-y^2}\ dy}_{-\frac{45}{8}\zeta(4)}-\frac13\int_0^1\frac{\ln^3(1+x)}{x}\ dx$$

For $\int_0^1\frac{\ln^3(1+x)}{x}\ dx$, we have a nice generalization here

$$\int_0^1\frac{\ln^n(1+x)}{x}\ dx=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)+\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$

$$\Longrightarrow \int_0^1\frac{\ln^3(1+x)}{x}\ dx=6\zeta(4)-\frac{21}{4}\ln2\zeta(3)+\frac32\ln^22\zeta(2)-\frac14\ln^42-6\operatorname{Li}_4\left(\frac12\right)$$

$$\Longrightarrow \mathcal{I}_2=\boxed{2\operatorname{Li}_4\left(\frac12\right)-\frac{5}{8}\zeta(4)+\frac{7}{4}\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42}$$

Combine the boxed results we get

$$ \mathcal{I}=-\operatorname{Li}_4\left(\frac12\right)+\frac{19}{16}\zeta(4)-\frac{1}{4}\ln2\zeta(3)-\frac14\ln^22\zeta(2)\frac1{24}\ln^42$$

Now substitute the results of $\mathcal{I}$ and $\mathcal{J}$ in $(1)$ we get

$$\sum_{n = 1}^\infty \frac{H_n \overline{H}_n}{n^2} = - 3 \operatorname{Li}_4 \left (\frac{1}{2} \right )+\frac{43}{16} \zeta (4) + \frac{3}{4} \ln^2 2\zeta (2)- \frac{1}{8} \ln^4 2$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.