1

How to prove

$$\sum_{n=1}^\infty\frac{(-1)^nH_{n/2}}{n^4}=\frac18\zeta(2)\zeta(3)-\frac{25}{32}\zeta(5)?$$

I came across this series while I was working on a nice integral $\int_0^1\frac{\ln(1+x)\operatorname{Li}_3(-x)}{x}dx$ and because I managed to calculate the integral in a different way, I got the closed form of the alternating series and solution will be posted soon.

Here is my question, is it possible to calculate the sum the same way @M.N.C.E calculated $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$ or by series manipulations? All approaches appreciated though. Thank you

By the way, is this result known in the literature?

Ali Shadhar
  • 25,498

2 Answers2

2

Using the identity

$$\int_0^1\frac{x^{2n}}{1+x}dx=\ln2+H_n-H_{2n}$$

Applying integration by parts yields

$$\int_0^1x^{2n-1}\ln(1+x)dx=\frac{H_{2n}-H_n}{2n}$$

Now replace $2n$ by $n$ then multiply both sides by $\frac{(-1)^n}{n^3}$ and sum up from $n=1$ and $\infty$ we obtain

$$\sum_{n=1}^\infty(-1)^n\frac{H_n-H_{n/2}}{n^4}=\int_0^1\frac{\ln(1+x)}{x}\sum_{n=1}^\infty \frac{(-x)^n}{n^3}dx$$

$$=\int_0^1\frac{\ln(1+x)}{x}\operatorname{Li}_3(-x)dx\overset{IBP}=-\frac{3}{8}\zeta(2)\zeta(3)+\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}dx\tag1$$

\begin{align} \int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx&=\int_0^1\frac1x\left(\frac12\operatorname{Li}_2(x^2)-\operatorname{Li}_2(x)\right)^2\ dx\\ &=\underbrace{\frac14\int_0^1\frac{\operatorname{Li}_2^2(x^2)}{x}\ dx}_{x^2=y}-\int_0^1\frac{\operatorname{Li}_2(x^2)\operatorname{Li}_2(x)}{x}\ dx+\int_0^1\frac{\operatorname{Li}_2^2(x)}{x}\ dx\\ &=\frac98\int_0^1\frac{\operatorname{Li}_2^2(x)}{x}\ dx-\int_0^1\frac{\operatorname{Li}_2(x^2)\operatorname{Li}_2(x)}{x}\ dx\\ &=\frac98\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx-\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^{2n-1}\operatorname{Li}_2(x)\ dx\\ &=\frac98\sum_{n=1}^\infty\frac1{n^2}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)-\sum_{n=1}^\infty\frac1{n^2}\left(\frac{\zeta(2)}{2n}-\frac{H_{2n}}{(2n)^2}\right)\\ &=\frac98\zeta(2)\zeta(3)-\frac98\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac12\zeta(2)\zeta(3)+4\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}\\ &=\frac58\zeta(2\zeta(3)+\frac78\sum_{n=1}^\infty\frac{H_n}{n^4}+2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}\tag{2} \end{align}

Now plug (2) in (1) we get

$$\sum_{n=1}^\infty\frac{(-1)^nH_{n/2}}{n^4}=-\frac14\zeta(2)\zeta(3)-\frac78\sum_{n=1}^\infty\frac{H_n}{n^4}-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$$

By substituting the following results:

$$\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$

$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$$

where the first sum can be calculated using Euler identity and the second one can be found here, the closed form of our series follows.

Ali Shadhar
  • 25,498
1

Different approach

$$S=\sum_{n=1}^\infty\frac{(-1)^nH_{n/2}}{n^4}=-H_{1/2}+\sum_{n=2}^\infty\frac{(-1)^nH_{n/2}}{n^4},\quad H_{1/2}=2\ln2-2$$

use the fact that

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

$$\Longrightarrow S=2-2\ln2+\frac1{16}\sum_{n=1}^\infty\frac{H_{n}}{n^4}-\sum_{n=1}^\infty\frac{H_{n+1/2}}{(2n+1)^4}$$

Lets compute the last sum,

notice that

$$H_{n+1/2}=2H_{2n+1}-H_n-2\ln2$$

$$\Longrightarrow \sum_{n=1}^\infty\frac{H_{n+1/2}}{n^4}=2\sum_{n=1}^\infty\frac{H_{2n+1}}{(2n+1)^4}-\sum_{n=1}^\infty\frac{H_{n}}{(2n+1)^4}-2\ln2\underbrace{\sum_{n=1}^\infty\frac{1}{(2n+1)^4}}_{\frac{15}{16}\zeta(4)-1}$$

where

\begin{align} 2\sum_{n=1}^\infty\frac{H_{2n+1}}{(2n+1)^4}&=2\sum_{n=0}^\infty\frac{H_{2n+1}}{(2n+1)^4}-2\\ &=\sum_{n=0}^\infty\frac{H_{n+1}}{(n+1)^4}+\sum_{n=0}^\infty\frac{(-1)^nH_{n+1}}{(n+1)^4}-2\\ &=\sum_{n=1}^\infty\frac{H_{n}}{n^4}-\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{n^4}-2 \end{align}

and

$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}=\frac{31}{8}\zeta(5)-\frac{15}8\ln2\zeta(4)-\frac{21}{16}\zeta(2)\zeta(3)$$

Which follows from using the generalization

$$\sum_{n=1}^\infty\frac{H_n}{(n+a)^2}=\left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2}$$

Combining the results gives

$$S=-\frac{15}{16}\sum_{n=1}^\infty\frac{H_n}{n^4}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}+\frac{31}{8}\zeta(5)-\frac{21}{16}\zeta(2)\zeta(3)$$

and by substituting the results of $\sum_{n=1}^\infty\frac{H_n}{n^4}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$ we get the claimed closed form.

Ali Shadhar
  • 25,498