Board index Computation of Series Some closed form of nonlinear Euler sums

Some closed form of nonlinear Euler sums

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Post Fri Mar 31, 2017 2:34 pm

Posts: 104
For a multi-index ${\bf S}=(s_1,s_2,\ldots,s_k)\ (k,s_i\in N, i=1,2,\ldots,k)$ with $s_1\leq s_2\leq \ldots\leq s_k$ and $q\geq 2$, the nonlinear Euler sums of index ${\bf S},q$ is defined by
$${S_{{\bf S},q}} := \sum\limits_{n = 1}^\infty {\frac{\zeta_n(s_1)\zeta_n(s_2)\cdots\zeta_n(s_k)}
{{{n^q}}}},$$
where the quantity $w:={s _1} + \cdots + {s _k} + q$ is called the weight, the quantity $k$ is called the degree.
As usual, repeated summands in partitions are indicated by powers, so that for instance
$${S_{{1^2}{2^3}4,q}} = {S_{112224,q}} = \sum\limits_{n = 1}^\infty {\frac{{H_n^2\zeta _n^3\left( 2 \right){\zeta _n}\left( 4 \right)}}{{{n^q}}}}. $$
Here $\zeta_n{(p)}$ stands for the generalized harmonic number defined by
$${\zeta _n}\left( p \right) := \sum\limits_{j = 1}^n {\frac{1}{{{j^p}}}},p>0,n\in N,$$
when $p=1$, ${H_n} := {\zeta _n}\left( 1 \right)$ is classical harmonic number, the empty sum $\zeta_0{(p)}$ is conventionally understood to be zero (Many papers use the notation $H^{(p)}_n$ to stands for the generalized harmonic number, namely $H^{(p)}_n\equiv \zeta_n{(p)}$).
In this paper ``Multiple zeta values and Euler sums" ([enter link description here][1]), the author obtain many closed form of Euler sums.
Some examples on nonlinear Euler sums follows:
\begin{align*}
&S_{1^22,2}=\sum\limits_{n = 1}^\infty {\frac{{H_n^2{\zeta _n}(2)}}{{{n^2}}}} = \frac{{41}}{{12}}\zeta \left( 6 \right) + 2{\zeta ^2}\left( 3 \right),\\
&S_{1^22,3}=\sum\limits_{n = 1}^\infty {\frac{{H_n^2{\zeta _n}\left( 2 \right)}}{{{n^3}}}} = - 7\zeta \left( 7 \right) + \frac{{19}}{2}\zeta \left( 3 \right)\zeta \left( 4 \right) - 2\zeta \left( 2 \right)\zeta \left( 5 \right),\\
&S_{1^32,3}=\sum\limits_{n =
1}^\infty {\frac{{H_n^3{\zeta _n}(2)}}{{{n^2}}}} =
\frac{{83}}{{16}}\zeta (7) + \frac{{27}}{2}\zeta (3)\zeta (4)- \frac{5}{2}\zeta (2)\zeta (5) ,\\
&S_{1^22,3}=\sum\limits_{n = 1}^\infty {\frac{{H_n^2{\zeta _n}(2)}}{{{n^3}}}}
= - 7\zeta (7) + \frac{{19}}{2}\zeta (3)\zeta (4) -
2\zeta (2)\zeta (5),\\
&S_{1^23,2}=\sum\limits_{n = 1}^\infty {\frac{{H_n^2{\zeta _n}(3)}}{{{n^2}}}}
= \frac{{329}}{{16}}\zeta (7)- 6\zeta (3)\zeta (4) -
\frac{9}{2}\zeta
(2)\zeta (5),\\
&S_{12^2,2}=\sum\limits_{n = 1}^\infty {\frac{{{H_n}\zeta
_n^2(2)}}{{{n^2}}}} = - \frac{{217}}{{16}}\zeta (7) + 5\zeta
(3)\zeta (4) + \frac{{13}}{2}\zeta (2)\zeta (5),\\
&S_{1^4,4}=\sum\limits_{n = 1}^\infty {\frac{{H_n^4}}{{{n^4}}}} = \frac{{13559}}{{144}}\zeta \left( 8 \right) - 92\zeta \left( 3 \right)\zeta \left( 5 \right) - 2\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) + 26{S_{2,6}},\\
&S_{1^22,4}=\sum\limits_{n = 1}^\infty {\frac{{H_n^2{\zeta _n}\left( 2 \right)}}{{{n^4}}}} = \frac{{193}}{{96}}\zeta \left( 8 \right) + 2\zeta \left( 3 \right)\zeta \left( 5 \right) - 2\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) + \frac{3}{2}{S_{2,6}},\\
&S_{1^24,2}=\sum\limits_{n = 1}^\infty {\frac{{H_n^2{\zeta _n}\left( 4 \right)}}{{{n^2}}}} = \frac{{1289}}{{96}}\zeta \left( 8 \right) - 11\zeta \left( 3 \right)\zeta \left( 5 \right) + 5{S_{2,6}},\\
&S_{1^23,3}=\sum\limits_{n = 1}^\infty {\frac{{H_n^2{\zeta _n}\left( 3 \right)}}{{{n^3}}}} = - \frac{{443}}{{288}}\zeta \left( 8 \right) + \frac{9}{2}\zeta \left( 3 \right)\zeta \left( 5 \right) + \frac{3}{2}\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) - \frac{{23}}{4}{S_{2,6}},\\
&{S_{{1^2}{2^2},2}} = \sum\limits_{n = 1}^\infty {\frac{{H_n^2\zeta _n^2\left( 2 \right)}}{{{n^2}}}} = \frac{{55}}{8}\zeta \left( 8 \right) - 7\zeta \left( 3 \right)\zeta \left( 5 \right) + 2\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) + 6{S_{2,6}},\\
&{S_{{3^2},2}} = \sum\limits_{n = 1}^\infty {\frac{{\zeta _n^2\left( 3 \right)}}{{{n^2}}}} = \frac{{677}}{{24}}\zeta \left( 8 \right) - 35\zeta \left( 3 \right)\zeta \left( 5 \right) + 4\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) + \frac{{15}}{2}{S_{2,6}},\\
&{S_{23,3}} = \sum\limits_{n = 1}^\infty {\frac{{{\zeta _n}\left( 2 \right){\zeta _n}\left( 3 \right)}}{{{n^3}}}} = - \frac{{827}}{{48}}\zeta \left( 8 \right)+\frac{{45}}{2}\zeta \left( 3 \right)\zeta \left( 5 \right) - \frac{3}{2}\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) - \frac{{23}}{4}{S_{2,6}},\\
&{S_{24,2}} = \sum\limits_{n = 1}^\infty {\frac{{\zeta _n^{}\left( 2 \right)\zeta _n^{}\left( 4 \right)}}{{{n^2}}}} = - \frac{{403}}{{36}}\zeta \left( 8 \right) + 20\zeta \left( 3 \right)\zeta \left( 5 \right) - 3\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) - \frac{9}{2}{S_{2,6}},\\
&S_{1^5,3}=\sum\limits_{n = 1}^\infty {\frac{{H_n^5}}{{{n^3}}}} = \frac{{60499}}{{288}}\zeta \left( 8 \right) - \frac{{393}}{2}\zeta \left( 3 \right)\zeta \left( 5 \right) - \frac{{{\rm{15}}}}{{\rm{2}}}\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) + \frac{{{\rm{235}}}}{{\rm{4}}}{S_{2,6}},\\
&{S_{{1^6},2}} = \sum\limits_{n = 1}^\infty {\frac{{H_n^6}}{{{n^2}}}} = \frac{{27903}}{6}\zeta \left( 8 \right) + 31\zeta \left( 3 \right)\zeta \left( 5 \right) + 16\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) + 57{S_{2,6}},\\
&S_{1^32,3}=\sum\limits_{n = 1}^\infty {\frac{{H_n^3{\zeta _n}\left( 2 \right)}}{{{n^3}}}} = - \frac{{2159}}{{48}}\zeta \left( 8 \right) + \frac{{93}}{2}\zeta \left( 3 \right)\zeta \left( 5 \right) + \frac{3}{2}\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) - \frac{{53}}{4}{S_{2,6}},\\
&S_{12^2,3}=\sum\limits_{n = 1}^\infty {\frac{{{H_n}\zeta _n^2\left( 2 \right)}}{{{n^3}}}} = - \frac{{6313}}{{288}}\zeta \left( 8 \right) + \frac{{43}}{2}\zeta \left( 3 \right)\zeta \left( 5 \right) + \frac{1}{2}\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) - \frac{{17}}{4}{S_{2,6}},\\
&{S_{{1^4}2,2}} = \sum\limits_{n = 1}^\infty {\frac{{H_n^4{\zeta _n}\left( 2 \right)}}{{{n^2}}}} = - \frac{{6631}}{{288}}\zeta \left( 8 \right) + 90\zeta \left( 3 \right)\zeta \left( 5 \right) + 3\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) - \frac{{47}}{2}{S_{2,6}},\\
&{S_{123,2}} = \sum\limits_{n = 1}^\infty {\frac{{{H_n}{\zeta _n}\left( 2 \right){\zeta _n}\left( 3 \right)}}{{{n^2}}}} = - \frac{{181}}{{288}}\zeta \left( 8 \right) + \frac{{15}}{2}\zeta \left( 3 \right)\zeta \left( 5 \right) - \frac{3}{2}\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) - \frac{7}{4}{S_{2,6}},\\
&{S_{{1^3}3,2}} = \sum\limits_{n = 1}^\infty {\frac{{H_n^3{\zeta _n}\left( 3 \right)}}{{{n^2}}}} = \frac{{809}}{{48}}\zeta \left( 8 \right) + \frac{{23}}{2}\zeta \left( 3 \right)\zeta \left( 5 \right) - \frac{7}{2}\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) - \frac{{33}}{4}{S_{2,6}},\\
&{S_{{1^5},4}} = \sum\limits_{n = 1}^\infty {\frac{{H_n^5}}{{{n^4}}}} = \frac{{4721}}{{36}}\zeta \left( 9 \right) + \frac{{265}}{8}\zeta \left( 2 \right)\zeta \left( 7 \right) - \frac{{4895}}{{24}}\zeta \left( 3 \right)\zeta \left( 6 \right) + 66\zeta \left( 4 \right)\zeta \left( 5 \right) - 5{\zeta ^3}\left( 3 \right),\\
&{S_{{1^2}3,4}} = \sum\limits_{n = 1}^\infty {\frac{{H_n^2{\zeta _n}\left( 3 \right)}}{{{n^4}}}} = \frac{{3895}}{{72}}\zeta \left( 9 \right) - \frac{5}{8}\zeta \left( 2 \right)\zeta \left( 7 \right) - \frac{{227}}{{24}}\zeta \left( 3 \right)\zeta \left( 6 \right) - \frac{{75}}{2}\zeta \left( 4 \right)\zeta \left( 5 \right) + {\zeta ^3}\left( 3 \right),\\
&{S_{{1^3}2,4}} = \sum\limits_{n = 1}^\infty {\frac{{H_n^3{\zeta _n}\left( 2 \right)}}{{{n^4}}}} = - \frac{{449}}{{36}}\zeta \left( 9 \right) - 7\zeta \left( 2 \right)\zeta \left( 7 \right) + \frac{{11}}{8}\zeta \left( 3 \right)\zeta \left( 6 \right) + 27\zeta \left( 4 \right)\zeta \left( 5 \right) - \frac{{11}}{3}{\zeta ^3}\left( 3 \right),\\
&{S_{{{12}^2},4}} = \sum\limits_{n = 1}^\infty {\frac{{{H_n}\zeta _n^2\left( 2 \right)}}{{{n^4}}}} = - \frac{{775}}{{36}}\zeta \left( 9 \right) + \frac{{85}}{8}\zeta \left( 2 \right)\zeta \left( 7 \right) - \frac{{221}}{{24}}\zeta \left( 3 \right)\zeta \left( 6 \right) + 10\zeta \left( 4 \right)\zeta \left( 5 \right) + 3{\zeta ^3}\left( 3 \right),\\
&S_{1^22,5}=\sum\limits_{n = 1}^\infty {\frac{{H_n^2{\zeta _n}\left( 2 \right)}}{{{n^5}}}} = - \frac{{1481}}{{72}}\zeta \left( 9 \right) - 3{\zeta ^3}\left( 3 \right) - 5\zeta \left( 2 \right)\zeta \left( 7 \right) + \frac{{295}}{{24}}\zeta \left( 3 \right)\zeta \left( 6 \right) + 18\zeta \left( 4 \right)\zeta \left( 5 \right),\\
&{S_{{{13}^2},3}} = \sum\limits_{n = 1}^\infty {\frac{{{H_n}\zeta _n^2\left( 3 \right)}}{{{n^3}}}} = \frac{{883}}{{20}}\zeta \left( {10} \right) - 26{\zeta ^2}\left( 5 \right) - \frac{{31}}{4}\zeta \left( 3 \right)\zeta \left( 7 \right) - 8\zeta \left( 2 \right)\zeta \left( 3 \right)\zeta \left( 5 \right)\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+ \frac{3}{4}{\zeta ^2}\left( 3 \right)\zeta \left( 4 \right) + 9\zeta \left( 2 \right){S_{2,6}} - \frac{{21}}{4}{S_{2,8}}.
\end{align*}


[1]: http://www.sciencedirect.com/science/ar ... 4X17300835

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