Series $\sum_{n=1}^{\infty} H_n\left[\zeta(8)-\frac{1}{1^8}-\frac{1}{2^8}-\frac{1}{3^8}\cdots\frac{1}{n^8}\right]$

Question

How can we find the sum of this hard series any hint please My trial is to express zeta(8) as a series but I don't know how and what to do

$$\sum_{n=1}^{\infty} H_n\left[\zeta(8)-\frac{1}{1^8}-\frac{1}{2^8}-\frac{1}{3^8}\cdots\frac{1}{n^8}\right]$$

Answer 1

By summation by parts

$$ S=\sum_{n=1}^{+\infty}H_n\left(\zeta(8)-H_n^{(8)}\right)= \sum_{n=1}^{+\infty}\left[(n+1)H_{n+1}-(n+1)\right]\frac{1}{(n+1)^8}\tag{1}$$ hence $$ S = -\zeta(7)+\sum_{n\geq 1}\frac{H_{n}}{n^7}=\color{red}{\frac{\pi^8}{4200}-\zeta(3)\zeta(5)-\zeta(7)}\tag{2} $$ by standard results on Euler sums.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\sum_{n = 1}^{\infty}H_{n}\bracks{\zeta\pars{8} - {1 \over 1^{8}} - {1 \over 2^{8}} - {1 \over 3^{8}} - \cdots - {1 \over n^{8}}} \\[5mm] = &\ \sum_{n = 1}^{\infty}H_{n}\pars{\zeta\pars{8} - \bracks{-\,{1 \over 7n^{7}} + \zeta\pars{8} + 8\int_{n}^{\infty}{\braces{x} \over x^{9}}\,\dd x}} \tag{$\large\color{#f00}{\S}$} \\[5mm] = &\ {1 \over 7}\ \underbrace{\sum_{n = 1}^{\infty}{H_{n} \over n^{7}}} _{\ds{{9 \over 2}\,\zeta\pars{8} - {1 \over 2}\sum_{n = 1}^{6}\zeta\pars{8 -n}\zeta\pars{n + 1}}}\ -\ 8\int_{1}^{\infty}{\sum_{n = 1}^{\infty}H_{n}\braces{nx}/n^{8} \over x^{9}} \,\dd x\tag{$\large\color{#f00}{\S\S}$} \\[1cm] = &\ {9 \over 14}\,\zeta\pars{8} - {1 \over 14}\,\zeta\pars{7}\zeta\pars{2} + {1 \over 7}\,\zeta\pars{6}\zeta\pars{3} + {1 \over 7}\,\zeta\pars{5}\zeta\pars{4} \\[2mm] - &\ 8\int_{1}^{\infty}{\sum_{n = 1}^{\infty}H_{n}\braces{nx}/n^{8} \over x^{9}} \,\dd x \end{align}

I'm still studying the integral. $\ds{\large\color{#f00}{\S}}$: See this expression. $\ds{\large\color{#f00}{\S\S}}$: See $\ds{\pars{24}}$ in a Harmonic Number page.

Answer 3

We shall begin by calculating the multivariate zeta value $\zeta(7,1)$ so we have it to hand later. Using "well known formula" and Euler reflection formula, we have \begin{eqnarray*} \zeta(8)&=&\zeta(7,1)+\zeta(6,2)+\zeta(5,3)+\zeta(4,4)+\zeta(3,5)+\zeta(2,6) \\ \zeta(2)\zeta(6)&=&\zeta(8)+\zeta(6,2)+\zeta(2,6) \\ \zeta(3)\zeta(5)&=&\zeta(8)+\zeta(5,3)+\zeta(3,5) \\ \zeta(4)^2&=&\zeta(8)+2\zeta(4,4). \\ \end{eqnarray*} Linear algebra gives $\zeta(7,1)=\frac{1}{2}(7\zeta_8-2\zeta_6\zeta_2 -2\zeta_5\zeta_3 -\zeta_4^2)$ & recall that $\zeta_2=\frac{\pi^2}{6},\zeta_4=\frac{\pi^4}{90},\zeta_6=\frac{\pi^6}{945},\zeta_8=\frac{\pi^8}{9450}$.

OK ... the question

\begin{eqnarray*} \sum_{n=1}^{\infty} H_n\left[\zeta(8)-\frac{1}{1^8}-\frac{1}{2^8}-\frac{1}{3^8}\cdots\frac{1}{n^8}\right] = \sum_{n=1}^{\infty} \sum_{m=n+1}^{\infty} \frac{H_n}{m^8} \\ = \sum_{n=1}^{\infty} \sum_{i=1}^{\infty} \frac{H_n}{(i+n)^8} = \sum_{n=1}^{\infty} \sum_{i=1}^{\infty} \sum_{j=1}^{n}\frac{1}{j(i+n)^8} \end{eqnarray*} Invert the order of the $j$ and $n$ plums, substitute $n=j+k-1$, \begin{eqnarray*} \sum_{i=1}^{\infty} \sum_{n=1}^{\infty} \sum_{j=1}^{n}\frac{1}{j(i+n)^8} =\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=j}^{\infty}\frac{1}{j(i+n)^8} =\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty}\frac{1}{j(i+j+k-1)^8} \end{eqnarray*} substitute $p=i+k-1$ & partial fractions $\frac{p}{j(j+p)}=\frac{1}{j}-\frac{1}{p+j}$ \begin{eqnarray*} \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty}\frac{1}{j(i+j+k-1)^8} =\sum_{p=1}^{\infty} \sum_{j=1}^{\infty}\frac{p}{j(p+j)^8} =\sum_{p=1}^{\infty} \sum_{j=1}^{n}\frac{1}{j(p+j)^7}-\sum_{p=1}^{\infty} \sum_{j=1}^{n}\frac{1}{(p+j)^8} \end{eqnarray*} note that the penultimate sum is $\zeta(7,1)$ & substitute $q=p+j$ in the last sum
\begin{eqnarray*} \sum_{p=1}^{\infty} \sum_{j=1}^{n}\frac{1}{(p+j)^8} = \sum_{q=1}^{\infty} \frac{q-1}{q^8} = \zeta(7)- \zeta(8). \end{eqnarray*} And so we have \begin{eqnarray*} \sum_{n=1}^{\infty} H_n\left[\zeta(8)-\frac{1}{1^8}-\frac{1}{2^8}-\frac{1}{3^8}\cdots\frac{1}{n^8}\right] = \color{red}{\frac{\pi^8}{4200}-\zeta(3)\zeta(5)-\zeta(7)}. \end{eqnarray*}