Linear Arrangement of Euler Sums with Multiple Argument

Abstract

We investigate the linear arrangement of Euler harmonic sums that may be expressed in closed form in terms of special functions such as the classical Riemann zeta function and the Dirichlet eta function. Particular emphasis is given to Euler harmonic sums with even weight. New examples highlighting the theorems will be presented.

1. Introduction and Background

Euler initiated the study of linear harmonic number sums of the form, using the notation of Flajolet and Salvy [1]

$${\mathbb{S}}_{p,t}^{++}(q):={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{H_{qn}^{(p)}}{n^t}},{\mathbb{S}}_{p,t}^{+-}(q):={\displaystyle \sum _{n=1}^{\infty }}{(-1)}^{n+1}{\displaystyle \frac{H_{qn}^{(p)}}{n^t}},$$

(1)

for the specific case $q=1$ and with the integer $p+t$ designated as the weight. The harmonic numbers $H_n^{\left(p\right)}$ of integer order p are defined as

$$H_n^{\left(p\right)}:={\displaystyle \sum _{r=1}^n}{\displaystyle \frac{1}{r^p}},p\in \mathbb{C}\mathrm{and}n\in \mathbb{N},$$

where $\mathbb{C}$ and $\mathbb{N}$ are the sets of complex numbers and positive integers, respectively. The harmonic numbers of order one, $H_{qn},$ are given by, for $q\in {\mathbb{R}}^{+}$ the set of positive real numbers,

$$\begin{array}{ccc}\hfill H_{qn}& =& \gamma +\psi \left(qn+1\right)\hfill \\ & =& \gamma +\psi \left(qn\right)+{\displaystyle \frac{1}{qn}}\left(n\in \mathbb{N}\right)\mathrm{and}{H}_0:=0.\hfill \end{array}$$

(2)

The term $\gamma$ represents the familiar Euler–Mascheroni constant (see, e.g., ([2], Section 1.2)), and $\psi \left(t\right)$ denotes the digamma (or psi) function defined by

$$\begin{array}{ccc}\hfill \psi \left(t\right)& :& ={\displaystyle \frac{d}{dt}}\left(log\Gamma \left(t\right)\right)={\displaystyle \frac{{\Gamma }^{\prime }\left(t\right)}{\Gamma \left(t\right)}}\left(t\in \mathbb{C}\setminus \left\{·,·,-3,-2,-1,0\right\}\right),\hfill \\ \hfill \mathrm{and}\psi \left(t+1\right)& =& \psi \left(t\right)+{\displaystyle \frac{1}{t}},\hfill \end{array}$$

where $\Gamma \left(t\right)$ is the familiar Gamma function (see, e.g., ([2], Section 1.1)). A generalization of the digamma function is the polygamma function ${\psi }^{\left(k\right)}(z)$ defined by

$$\begin{array}{ccc}\hfill {\psi }^{\left(k\right)}(z)& :& ={\displaystyle \frac{d^k}{d{z}^k}}\left\{\psi (z)\right\}={\left(-1\right)}^{k+1}k!{\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{1}{{\left(r+z\right)}^{k+1}}}={\left(-1\right)}^{k+1}k!\zeta (k+1,z),\hfill \\ & & \left(k\in \mathbb{N};z\in \mathbb{C}\setminus \left\{·,·,-3,-2,-1,0\right\}\right).\hfill \end{array}$$

(3)

In his works on series, Euler solved the famous Basel problem,

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n^2}}=\zeta \left(2\right)={\displaystyle \frac{{\pi }^2}{6}}.$$

Furthermore, Euler proved from the properties of the Bernoulli numbers, $B_0=1$, $B_1=-{\displaystyle \frac{1}{2}}$, $B_2={\displaystyle \frac{1}{6}},B_{2r+1}=0,$ for $r\ge 1,$ given by the generating function

$${\displaystyle \frac{w}{exp\left(w\right)-1}}=\sum _{j\ge 0}{\displaystyle \frac{w^j}{j!}}Bj,\mathrm{for}\left|w\right|<2\pi$$

and the classical Euler–Bernoulli relation (see [2], p. 166), that

$$\zeta \left(2r\right)={\displaystyle \frac{{\left(-1\right)}^{r+1}{B}_{2r}{2}^{2r-1}{\pi }^{2r}}{\left(2r\right)!}}.$$

The Riemann zeta function, $\zeta \left(t\right)$, is defined as

$$\zeta \left(t\right):=\left\{\begin{array}{c}\sum _{j\ge 1}{\displaystyle \frac{1}{j^t}}={\displaystyle \frac{1}{1-2^{-t}}}\sum _{j\ge 1}{\displaystyle \frac{1}{{\left(2j-1\right)}^t}};\Re (t)>1\\ \\ {\displaystyle \frac{1}{1-2^{1-t}}}\sum _{j\ge 1}{\displaystyle \frac{{\left(-1\right)}^{j-1}}{j^t}},\Re (t)>0,t\ne 1\end{array}\right.,$$

The Dirichlet eta function, $\eta (t)$ is defined by

$$\eta (t):=\sum _{n⩾1}{\displaystyle \frac{{(-1)}^{n-1}}{n^t}}=\left(1-2^{1-t}\right)\zeta (t)(\Re (t)>0),$$

where $\Re (t)$ defines the real part of t. The objective of this paper is to obtain linear arrangements of (1) in closed form in terms of special functions, such as

$$\begin{array}{ccc}& & {\displaystyle \frac{{\left(-1\right)}^{t}t}{q^{t+1}}}\left({\mathbb{S}}_{1,t+2}^{++}(q)-{\mathbb{S}}_{1,t+2}^{++}({\displaystyle \frac{q}{2}})+{\displaystyle \frac{{\left(-1\right)}^t}{q^{-t-1}t}}{\mathbb{S}}_{1,t+2}^{+-}({\displaystyle \frac{1}{q}})\right)\hfill \\ & & +{\displaystyle \frac{{\left(-1\right)}^t}{q^t}}\left({\mathbb{S}}_{2,t+1}^{++}(q)-{\displaystyle \frac{1}{2}}{\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{q}{2}})+{\displaystyle \frac{{\left(-1\right)}^t}{q^{1-t}}}{\mathbb{S}}_{2,t+1}^{+-}({\displaystyle \frac{1}{q}})\right)\hfill \end{array}$$

$$={\displaystyle \frac{1}{q}}\zeta \left(2\right)\eta \left(t+1\right)+{\displaystyle \frac{{\left(-1\right)}^t}{q^t}}\eta \left(2\right)\zeta \left(t+1\right)+{\left(-1\right)}^t{\displaystyle \sum _{j=2}^t}{\displaystyle \frac{{\left(-1\right)}^j}{q^{t+2-j}}}\left(t+1-j\right)\eta \left(j\right)\zeta (t+3-j),$$

(4)

and, in a specific case, a linear arrangement of harmonic sums with even weight

$$\begin{array}{ccc}& & 2{\mathbb{S}}_{2,12}^{+-}-{\displaystyle \frac{4095}{4096}}{\mathbb{S}}_{2,12}^{++}-{\mathbb{S}}_{12,2}^{+-}+12{\mathbb{S}}_{1,13}^{+-}\hfill \\ & =& {\displaystyle \frac{75,989,887}{1,720,320}}\zeta (14)-{\displaystyle \frac{189}{32}}\zeta {(7)}^2-{\displaystyle \frac{375}{32}}\zeta (5)\zeta (9)-{\displaystyle \frac{5883}{512}}\zeta (11)\zeta (3).\hfill \end{array}$$

To the author’s knowledge, there are few, if any, expressions of the type (4) in the published literature. The idea of expressing infinite series, of Euler type, in closed form is often difficult but often important. Any expression represented in closed form has obvious and desirable properties, as may be ascertained from a published work of Borwein and Crandall [3]. It is generally accepted that ${\mathbb{S}}_{p,q}^{++}$ may not have a closed-form representation other than the pair $\left\{\left(2,4\right),\left(4,2\right)\right\}$ and $p=q.$ If ${\mathbb{S}}_{p,q}^{++}$ can be evaluated, then ${\mathbb{S}}_{q,p}^{++}$ exists by virtue of the shuffle relation

$${\mathbb{S}}_{p,q}^{++}+{\mathbb{S}}_{q,p}^{++}=\zeta (p)\zeta (q)+\zeta (p+q).$$

The symmetrical nature of the shuffle relation and the identities established in this paper lend themselves to further exploration. In a forthcoming paper, we shall exploit the identity (4) in another symmetrical arrangement of harmonic numbers of a higher order. For the case $q=1$, we recall the following known results

$${\mathbb{S}}_{1,t}^{++}=\left(1+{\displaystyle \frac{t}{2}}\right)\zeta \left(t+1\right)-{\displaystyle \frac{1}{2}}{\displaystyle \sum _{j=1}^{t-2}}\zeta \left(j+1\right)\zeta \left(t-j\right),$$

(5)

due to Euler [4]. For odd weight $p+t$, Borwein et al. [5] gave the identity:

$${\mathbb{S}}_{p,t}^{++}={\displaystyle \frac{1-{\left(-1\right)}^p}{2}}\zeta \left(p\right)\zeta \left(t\right)+\left(1-{\left(-1\right)}^p\left({\displaystyle \frac{p+t}{t}}\right)\left(\begin{array}{c}p+t-1\\ p\end{array}\right)\right){\displaystyle \frac{\zeta \left(p+t\right)}{2}}$$

(6)

$$+{\left(-1\right)}^p{\displaystyle \sum _{j=1}^{\left[{\displaystyle \frac{p}{2}}\right]}}\left(\begin{array}{c}p+t-2j-1\\ t-1\end{array}\right)\zeta \left(2j\right)\zeta \left(p+t-2j\right)$$

$$+{\left(-1\right)}^p{\displaystyle \sum _{j=1}^{\left[{\displaystyle \frac{t}{2}}\right]}}\left(\begin{array}{c}p+t-2j-1\\ p-1\end{array}\right)\zeta \left(2j\right)\zeta \left(p+t-2j\right).$$

For the alternating case, for odd weight $1+t$, Sitaramachandrarao [6] published

$$2{\mathbb{S}}_{1,t}^{+-}=\left(t+1\right)\eta \left(t+1\right)-\zeta \left(t+1\right)-2{\displaystyle \sum _{j=1}^{{\displaystyle \frac{t}{2}}-1}}\zeta \left(2j\right)\zeta \left(t+1-2j\right).$$

(7)

Flajolet and Salvy [1] gave the expression for odd weight $p+t$

$$2{\mathbb{S}}_{p,t}^{+-}=\left(1-{\left(-1\right)}^p\right)\zeta \left(p\right)\eta \left(t\right)+\eta \left(p+t\right)$$

(8)

$$\begin{array}{ccc}& & +2\sum _{j+2k=p}\left(\begin{array}{c}t+j-1\\ t-1\end{array}\right){\left(-1\right)}^{j+1}\eta \left(t+j\right)\eta \left(2k\right)\hfill \\ & & +2{\left(-1\right)}^p\sum _{i+2k=t}\left(\begin{array}{c}p+i-1\\ p-1\end{array}\right)\zeta \left(p+i\right)\eta \left(2k\right),\hfill \end{array}$$

where $\eta \left(·\right)$ is the Dirichlet eta function. Recently, Alzer and Choi [7] obtained a good result for $p\in \mathbb{N}\setminus \left\{1\right\}$

$$2{\mathbb{S}}_{p,1}^{+-}=p\zeta \left(p+1\right)-\zeta \left(t+1\right)-2{\displaystyle \sum _{j=1}^p}\eta \left(j\right)\eta \left(p+1-j\right).$$

(9)

The polylogarithm function ${\mathrm{Li}}_p(z)$ of order $p,$ and for each integer $p\ge 1,$ is defined by (see, e.g., [2], p. 198)

$${\mathrm{Li}}_p(z)={\displaystyle \sum _{m=1}^{\infty }}{\displaystyle \frac{z^m}{m^p}}\left(\left|z\right|⩽1;p\in \mathbb{N}\ge 2\right)$$

(10)

and

$${\mathrm{Li}}_1(z)=-log\left(1-z\right),\left(\left|z\right|⩽1\right).$$

There are many interesting and significant results associated with Euler harmonic sum identities, some of which may be seen in the works of [5,8,9,10,11,12,13,14]. Li and Chu [15] have published some important results in relation to generating functions for binomial series involving harmonic-like numbers, and Chen [16] has investigated some general Tornheim-type series. The majority of the published works that deal with Euler harmonic sums of the type (11) deal with the case $q=1.$

This paper will be organized in the following way. The main results will be established, proving that

$${\displaystyle \frac{{\left(-1\right)}^t}{q^t}}\left(1+t\right){\mathbb{S}}_{1,t+2}^{++}(q)+{\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{1}{q}})+{\displaystyle \frac{{\left(-1\right)}^t}{q^{t-1}}}{\mathbb{S}}_{2,t+1}^{++}(q)$$

has a closed form. A special case in the form of a Corollary is detailed. The second theorem deals with expressing the linear arrangement

$${\displaystyle \frac{{\left(-1\right)}^{t}t}{q^{t+1}}}\left({\mathbb{S}}_{1,t+2}^{++}(q)-{\mathbb{S}}_{1,t+2}^{++}({\displaystyle \frac{q}{2}})\right)+{\mathbb{S}}_{1,t+2}^{+-}({\displaystyle \frac{1}{q}})+{\displaystyle \frac{{\left(-1\right)}^t}{q^t}}\left({\mathbb{S}}_{2,t+1}^{++}(q)-{\displaystyle \frac{1}{2}}{\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{q}{2}})\right)+{\displaystyle \frac{1}{q}}{\mathbb{S}}_{2,t+1}^{+-}({\displaystyle \frac{1}{q}})$$

in closed form with its resultant corollary. Some examples will be given, highlighting the linear arrangement of Euler sums with even weight in closed form.

2. The Main Results

Consider the following theorem.

Theorem 1.

Let $t\in \mathbb{N},t\ge 2$ and let $q\in {\mathbb{R}}^{+}\setminus \left\{0\right\}$; the following identity is valid:

$${\displaystyle \frac{{\left(-1\right)}^t}{q^t}}\left(1+t\right){\mathbb{S}}_{1,t+2}^{++}(q)+{\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{1}{q}})+{\displaystyle \frac{{\left(-1\right)}^t}{q^{t-1}}}{\mathbb{S}}_{2,t+1}^{++}(q)$$

(11)

$$={\displaystyle \frac{1}{q}}\zeta (3)\zeta (t)+{\left(-1\right)}^{t}t{q}^{1-t}\zeta (2)\zeta (t+1)$$

$$+{\displaystyle \sum _{j=3}^t}{\left(-1\right)}^j\left({\displaystyle \frac{{\left(-1\right)}^t}{q^{t+1-j}}}\left(t+1-j\right)\zeta (j)\zeta \left(t+3-j\right)+{\displaystyle \frac{1}{q^{j-1}}}\zeta (j+1)\zeta \left(t+2-j\right)\right),$$

where ${\mathbb{S}}_{1,t+2}^{++}(q),{\mathbb{S}}_{2,t+1}^{++}(q)$ are the linear Euler harmonic sum (1) of weight $t+3$, and $\zeta \left(t\right)$ is the classical Riemann zeta function.

Proof. 

Consider the standard integral representation for the harmonic number

$$-{\displaystyle \frac{q}{n}}{H}_{{\displaystyle \frac{n}{q}}}:={\int }_0^1{x}^{{\displaystyle \frac{n}{q}}-1}ln\left(1-x\right)\mathrm{d}x={\int }_0^1{\displaystyle \frac{ln\left(1-x\right)}{x}}{x}^{{\displaystyle \frac{n}{q}}}\mathrm{d}x,$$

(12)

Then,

$${\displaystyle \frac{\partial }{\partial q}}\left(-{\displaystyle \frac{q}{n}}{H}_{{\displaystyle \frac{n}{q}}}\right)={\int }_0^1{\displaystyle \frac{ln\left(1-x\right)}{x}}{\displaystyle \frac{\partial }{\partial q}}\left(x^{{\displaystyle \frac{n}{q}}}\right)\mathrm{d}x$$

$${\displaystyle \frac{q^2}{n^2}}{H}_{{\displaystyle \frac{n}{q}}}+{\displaystyle \frac{q}{n}}{H}_{{\displaystyle \frac{n}{q}}}^{\left(2\right)}-q\zeta \left(2\right)={\int }_0^1{\displaystyle \frac{ln\left(1-x\right)lnx}{x}}{x}^{{\displaystyle \frac{n}{q}}}\mathrm{d}x.$$

Summing over the integers n for $t\in \mathbb{N},t\ge 2,$

$$\sum _{n\ge 1}{\displaystyle \frac{1}{n^t}}\left({\displaystyle \frac{q^2}{n^2}}{H}_{{\displaystyle \frac{n}{q}}}+{\displaystyle \frac{q}{n}}{H}_{{\displaystyle \frac{n}{q}}}^{\left(2\right)}-{\displaystyle \frac{q}{n}}\zeta \left(2\right)\right)={\int }_0^1{\displaystyle \frac{lnxln\left(1-x\right)}{x}}\sum _{n\ge 1}{\displaystyle \frac{x^{\frac{n}{q}}}{n^t}}\mathrm{d}x,$$

and applying definition (10), we obtain

$$\sum _{n\ge 1}{\displaystyle \frac{1}{n^t}}\left({\displaystyle \frac{q^2}{n^2}}{H}_{{\displaystyle \frac{n}{q}}}+{\displaystyle \frac{q}{n}}{H}_{{\displaystyle \frac{n}{q}}}^{\left(2\right)}-{\displaystyle \frac{q}{n}}\zeta \left(2\right)\right)={\int }_0^1{\displaystyle \frac{lnxln\left(1-x\right)}{x}}{\mathrm{Li}}_t(x^{{\displaystyle \frac{1}{q}}})\mathrm{d}x.$$

Using the transformation $y^q=x$, we obtain

$${\mathbb{S}}_{1,t+2}^{++}({\displaystyle \frac{1}{q}})+{\displaystyle \frac{1}{q}}{\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{1}{q}})-{\displaystyle \frac{1}{q}}\zeta \left(2\right)\zeta \left(t+1\right)={\int }_0^1{\displaystyle \frac{lnxln\left(1-y^q\right)}{y}}{\mathrm{Li}}_t(y)\mathrm{d}y,$$

A Taylor series expansion of the $ln\left(1-y^q\right)$ term gives us the representation

$${\mathbb{S}}_{1,t+2}^{++}({\displaystyle \frac{1}{q}})+{\displaystyle \frac{1}{q}}{\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{1}{q}})-{\displaystyle \frac{1}{q}}\zeta \left(2\right)\zeta \left(t+1\right)=-\sum _{n\ge 1}{\displaystyle \frac{1}{n}}{\displaystyle \underset{0}{\overset{1}{\int }}}{y}^{qn-1}lnx{\mathrm{Li}}_t(y)\mathrm{d}y$$

$$\begin{array}{ccc}& =& \sum _{n\ge 1}{\displaystyle \frac{t{\left(-1\right)}^{t+1}{H}_{nq}}{n{\left(nq\right)}^{t+1}}}+\sum _{n\ge 1}{\displaystyle \frac{{\left(-1\right)}^{t+1}\left(\zeta (2)-H_{nq}^{\left(2\right)}\right)}{n{\left(nq\right)}^t}}\hfill \\ & & +{\left(-1\right)}^t{\displaystyle \sum _{j=2}^t}{\left(-1\right)}^j\left(t+1-j\right)\zeta (j)\sum _{n\ge 1}{\displaystyle \frac{1}{n{\left(nq\right)}^{t+2-j}}}\hfill \end{array}$$

$$\begin{array}{ccc}& =& {\displaystyle \frac{t{\left(-1\right)}^{t+1}}{q^{t+1}}}{\mathbb{S}}_{1,t+2}^{++}(q)+{\displaystyle \frac{{\left(-1\right)}^t}{q^t}}\zeta \left(2\right)\zeta \left(t+1\right)+{\displaystyle \frac{{\left(-1\right)}^t}{q^t}}{\mathbb{S}}_{2,t+1}^{++}(q)\hfill \\ & & +{\left(-1\right)}^t{\displaystyle \sum _{j=2}^t}{\displaystyle \frac{{\left(-1\right)}^j}{q^{t+2-j}}}\left(t+1-j\right)\zeta (j)\zeta (t+3-j)\hfill \end{array}$$

$$={\displaystyle \frac{t{\left(-1\right)}^{t+1}}{q^{t+1}}}{\mathbb{S}}_{1,t+2}^{++}(q)+{\displaystyle \frac{{\left(-1\right)}^t}{q^t}}{\mathbb{S}}_{2,t+1}^{++}(q)+{\left(-1\right)}^t{\displaystyle \sum _{j=3}^t}{\displaystyle \frac{{\left(-1\right)}^j}{q^{t+2-j}}}\left(t+1-j\right)\zeta (j)\zeta (t+3-j).$$

By rearranging, we obtain

$$\begin{array}{ccc}& & q{\mathbb{S}}_{1,t+2}^{++}({\displaystyle \frac{1}{q}})+{\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{1}{q}})+{\displaystyle \frac{t{\left(-1\right)}^t}{q^t}}{\mathbb{S}}_{1,t+2}^{++}(q)+{\displaystyle \frac{{\left(-1\right)}^t}{q^{t-1}}}{\mathbb{S}}_{2,t+1}^{++}(q)\hfill \\ & =& \left(1+{\displaystyle \frac{{\left(-1\right)}^{t}t}{q^{t-1}}}\right)\zeta \left(2\right)\zeta \left(t+1\right)+{\left(-1\right)}^t{\displaystyle \sum _{j=3}^t}{\displaystyle \frac{{\left(-1\right)}^j}{q^{t+1-j}}}\left(t+1-j\right)\zeta (j)\zeta (t+3-j),\hfill \end{array}$$

From the following relation (see [11]), we substitute the identity

$${\mathbb{S}}_{1,t+1}^{++}({\displaystyle \frac{1}{q}})={\displaystyle \frac{{\left(-1\right)}^{t+1}}{q^t}}{\mathbb{S}}_{1,t+1}^{++}(q)+{\displaystyle \sum _{j=1}^{t-1}}{\displaystyle \frac{{\left(-1\right)}^{j+1}}{q^j}}\zeta (j+1)\zeta \left(t+1-j\right)$$

into the previous equation; hence,

$${\displaystyle \frac{{\left(-1\right)}^t}{q^t}}\left(1+t\right){\mathbb{S}}_{1,t+2}^{++}(q)+{\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{1}{q}})+{\displaystyle \frac{{\left(-1\right)}^t}{q^{t-1}}}{\mathbb{S}}_{2,t+1}^{++}(q)=\left(1+{\left(-1\right)}^{t}t{q}^{1-t}\right)\zeta (2)\zeta (t+1)$$

$$+{\left(-1\right)}^t{\displaystyle \sum _{j=3}^t}{\displaystyle \frac{{\left(-1\right)}^j}{q^{t+1-j}}}\left(t+1-j\right)\zeta (j)\zeta \left(t+3-j\right)+{\displaystyle \sum _{j=1}^t}{\displaystyle \frac{{\left(-1\right)}^j}{q^{j-1}}}\zeta (j+1)\zeta \left(t+2-j\right).$$

Here, we isolate the $j=1,2$ terms of the second sum and the proof of the theorem is complete. □

As a direct consequence of Theorem 1, we can highlight some special cases, as in the following corollary.

Corollary 1.

From Theorem 1, we can highlight some special cases. Consider the case of $q=1$; then,

$${\left(-1\right)}^t\left(1+t\right){\mathbb{S}}_{1,t+2}^{++}+{\mathbb{S}}_{2,t+1}^{++}+{\left(-1\right)}^t{\mathbb{S}}_{2,t+1}^{++}={\left(-1\right)}^{t}t\zeta (2)\zeta (t+1)+\zeta (3)\zeta (t)$$

$$+{\displaystyle \sum _{j=3}^t}{\left(-1\right)}^j\left({\left(-1\right)}^t\left(t+1-j\right)\zeta (j)\zeta \left(t+3-j\right)+\zeta (j+1)\zeta \left(t+2-j\right)\right).$$

If t is even, let $t=2t^{*}$ (and rename $t^{*}=t$); hence,

$$\begin{array}{ccc}\hfill 2{\mathbb{S}}_{2,2t+1}^{++}& =& 2t\zeta (2)\zeta (2t+1)+\zeta (3)\zeta (2t)-\left(1+2t\right){\mathbb{S}}_{1,2t+2}^{++}\hfill \\ & & +{\displaystyle \sum _{j=3}^{2t}}{\left(-1\right)}^j\left(\left(2t+1-j\right)\zeta (j)\zeta \left(2t+3-j\right)+\zeta (j+1)\zeta \left(2t+2-j\right)\right).\hfill \end{array}$$

If t is odd, let $t=2t^{*}-1$ (and rename $t^{*}=t$); hence,

$$\begin{array}{ccc}\hfill 2t{\mathbb{S}}_{1,2t+1}^{++}& =& \left(2t-1\right)\zeta (2)\zeta (2t)+\zeta (3)\zeta (2t-1)\hfill \\ & & +{\displaystyle \sum _{j=3}^{2t}}{\left(-1\right)}^j\left(\zeta (j+1)\zeta \left(2t+1-j\right)-\left(2t-j\right)\zeta (j)\zeta \left(2t+2-j\right)\right).\hfill \end{array}$$

From (11), for the case $q=2$, let $t=2t^{*}-1$ (and rename $t^{*}=t$), where $t\in \mathbb{N}$, we obtain

$$-{\displaystyle \frac{2t}{2^{2t-1}}}{\mathbb{S}}_{1,2t+1}^{++}(2)+{\mathbb{S}}_{2,2t}^{++}({\displaystyle \frac{1}{2}})-{\displaystyle \frac{1}{2^{2t-2}}}{\mathbb{S}}_{2,2t}^{++}(2)$$

(13)

$$={\displaystyle \frac{1}{2}}\zeta (3)\zeta (2t-1)-{\displaystyle \frac{\left(2t-1\right)}{2^{2t-2}}}\zeta (2)\zeta (2t)$$

$$+{\displaystyle \sum _{j=3}^{2t-1}}{\left(-1\right)}^j\left({\displaystyle \frac{1}{2^{j-1}}}\zeta (j+1)\zeta \left(2t+1-j\right)-{\displaystyle \frac{\left(2t-j\right)}{2^{2t-j}}}\zeta (j)\zeta \left(2t+2-j\right)\right).$$

By applying the decomposition relation

$$2^{-t}{\mathbb{S}}_{1,t+1}^{++}(q)={\mathbb{S}}_{1,t+1}^{++}\left({\displaystyle \frac{q}{2}}\right)-{\mathbb{S}}_{1,t+1}^{+-}\left({\displaystyle \frac{q}{2}}\right),$$

(14)

we can rewrite (13) as

$$4t{\mathbb{S}}_{1,2t+1}^{+-}+{\mathbb{S}}_{2,2t}^{++}({\displaystyle \frac{1}{2}})+2{\mathbb{S}}_{2,2t}^{+-}-2{\mathbb{S}}_{2,2t}^{++}$$

(15)

$$=4t{\mathbb{S}}_{1,2t+1}^{++}+{\displaystyle \frac{1}{2}}\zeta (3)\zeta (2t-1)-{\displaystyle \frac{\left(2t-1\right)}{2^{2t-2}}}\zeta (2)\zeta (2t)$$

$$+{\displaystyle \sum _{j=3}^{2t-1}}{\left(-1\right)}^j\left({\displaystyle \frac{1}{2^{j-1}}}\zeta (j+1)\zeta \left(2t+1-j\right)-{\displaystyle \frac{\left(2t-j\right)}{2^{2t-j}}}\zeta (j)\zeta \left(2t+2-j\right)\right),$$

where ${\mathbb{S}}_{1,2t+1}^{++}$ is the Euler identity (5). In ref. [17], the authors proved that

$${\mathbb{S}}_{2,2t}^{++}({\displaystyle \frac{1}{2}})=2^{1-2t}{\mathbb{S}}_{2,2t}^{++}+2{\mathbb{S}}_{2,2t}^{+-}-2{\mathbb{S}}_{2t,2}^{+-}-2\eta \left(2\right)\eta \left(2t\right)+2\eta \left(2t+2\right)$$

By simplifying (15), we have that

$$4t{\mathbb{S}}_{1,2t+1}^{+-}+\left(2^{1-2t}-2\right){\mathbb{S}}_{2,2t}^{++}+4{\mathbb{S}}_{2,2t}^{+-}-2{\mathbb{S}}_{2t,2}^{+-}=2\eta \left(2\right)\eta \left(2t\right)-2\eta \left(2t+2\right)$$

(16)

$$+4t{\mathbb{S}}_{1,2t+1}^{++}+{\displaystyle \frac{1}{2}}\zeta (3)\zeta (2t-1)-{\displaystyle \frac{\left(2t-1\right)}{2^{2t-2}}}\zeta (2)\zeta (2t)$$

$$+{\displaystyle \sum _{j=3}^{2t-1}}{\left(-1\right)}^j\left({\displaystyle \frac{1}{2^{j-1}}}\zeta (j+1)\zeta \left(2t+1-j\right)-{\displaystyle \frac{\left(2t-j\right)}{2^{2t-j}}}\zeta (j)\zeta \left(2t+2-j\right)\right).$$

The individual Euler sums in (15) and (16) are not known explicitly in closed form for $t\ge 3.$ Therefore, (15) and (16) provide a linear arrangement of Euler sums with even weight that can be expressed in closed form.

For the case $t=4$, we have

$$16{\mathbb{S}}_{1,9}^{+-}+{\mathbb{S}}_{2,8}^{++}({\displaystyle \frac{1}{2}})+2{\mathbb{S}}_{2,8}^{+-}-2{\mathbb{S}}_{2,8}^{++}={\displaystyle \frac{341}{8}}\zeta (10)-{\displaystyle \frac{15}{2}}\zeta {(5)}^2-{\displaystyle \frac{237}{16}}\zeta (7)\zeta (3),$$

and

$$16{\mathbb{S}}_{1,9}^{+-}-{\displaystyle \frac{255}{128}}{\mathbb{S}}_{2,8}^{++}+4{\mathbb{S}}_{2,8}^{+-}-2{\mathbb{S}}_{8,2}^{+-}={\displaystyle \frac{108201}{2560}}\zeta (10)-{\displaystyle \frac{15}{2}}\zeta {(5)}^2-{\displaystyle \frac{237}{16}}\zeta (7)\zeta (3).$$

(17)

For the case where t is even, from (12), let $t=2t^{*}$ (and rename $t^{*}=t$), where $t\in \mathbb{N},$

$${\displaystyle \frac{1}{2^{2t}}}\left(1+2t\right){\mathbb{S}}_{1,2t+2}^{++}(2)+{\mathbb{S}}_{2,2t+1}^{++}({\displaystyle \frac{1}{2}})+{\displaystyle \frac{1}{2^{2t-1}}}{\mathbb{S}}_{2,2t+1}^{++}(2)$$

$$={\displaystyle \frac{1}{2}}\zeta (3)\zeta (2t)+t{2}^{2-2t}\zeta (2)\zeta (2t+1)$$

$$+{\displaystyle \sum _{j=3}^{2t}}{\left(-1\right)}^j\left({\displaystyle \frac{1}{2^{2t+1-j}}}\left(2t+1-j\right)\zeta (j)\zeta \left(2t+3-j\right)+{\displaystyle \frac{1}{2^{j-1}}}\zeta (j+1)\zeta \left(2t+2-j\right)\right).$$

From ref. [17] (p. 414), we have that

$${\mathbb{S}}_{2,2t1}^{++}({\displaystyle \frac{1}{2}})=2^{-2t}{\mathbb{S}}_{2,2t+1}^{++}-22{\mathbb{S}}_{2,2t}^{+-}-2{\mathbb{S}}_{2t+1,2}^{+-}-2\eta \left(2\right)\eta \left(2t+1\right)+2\eta \left(2t+3\right)$$

and, therefore, an explicit representation of the linear arrangement of

$${\displaystyle \frac{1}{2^{2t}}}\left(1+2t\right){\mathbb{S}}_{1,2t+2}^{++}(2)+{\displaystyle \frac{1}{2^{2t-1}}}{\mathbb{S}}_{2,2t+1}^{++}(2)={\displaystyle \frac{1}{2}}\zeta (3)\zeta (2t)+t{2}^{2-2t}\zeta (2)\zeta (2t+1)$$

$$-{\mathbb{S}}_{2,2t+1}^{++}({\displaystyle \frac{1}{2}})+{\displaystyle \sum _{j=3}^{2t}}{\left(-1\right)}^j\left({\displaystyle \frac{1}{2^{2t+1-j}}}\left(2t+1-j\right)\zeta (j)\zeta \left(2t+3-j\right)+{\displaystyle \frac{1}{2^{j-1}}}\zeta (j+1)\zeta \left(2t+2-j\right)\right).$$

In the next theorem, we establish a linear arrangement of Euler sums involving the alternating sum ${\mathbb{S}}_{2,t+1}^{+-}({\displaystyle \frac{1}{q}})$, which was not evident in Theorem 1.

Theorem 2.

Let $t\in \mathbb{N},t\ge 2$ and let  $q\in {\mathbb{R}}^{+}\setminus \left\{0\right\}$ the following identity is valid:

$$\begin{array}{ccc}& & {\displaystyle \frac{{\left(-1\right)}^{t}t}{q^{t+1}}}\left({\mathbb{S}}_{1,t+2}^{++}(q)-{\mathbb{S}}_{1,t+2}^{++}({\displaystyle \frac{q}{2}})\right)+{\mathbb{S}}_{1,t+2}^{+-}({\displaystyle \frac{1}{q}})\hfill \\ & & +{\displaystyle \frac{{\left(-1\right)}^t}{q^t}}\left({\mathbb{S}}_{2,t+1}^{++}(q)-{\displaystyle \frac{1}{2}}{\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{q}{2}})\right)+{\displaystyle \frac{1}{q}}{\mathbb{S}}_{2,t+1}^{+-}({\displaystyle \frac{1}{q}})\hfill \end{array}$$

(18)

$$={\displaystyle \frac{1}{q}}\zeta \left(2\right)\eta \left(t+1\right)+{\displaystyle \frac{{\left(-1\right)}^t}{q^t}}\eta \left(2\right)\zeta \left(t+1\right)+{\left(-1\right)}^t{\displaystyle \sum _{j=2}^t}{\displaystyle \frac{{\left(-1\right)}^j}{q^{t+2-j}}}\left(t+1-j\right)\eta \left(j\right)\zeta (t+3-j),$$

where ${\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{q}{2}})$ is the linear Euler harmonic sum (1) of weight $t+3$, and $\eta \left(t\right)$ is the classical Riemann eta function.

Proof. 

As in Theorem 1, the standard integral representation for the harmonic number

$$-{\displaystyle \frac{q}{n}}{H}_{{\displaystyle \frac{n}{q}}}:={\int }_0^1{x}^{{\displaystyle \frac{n}{q}}-1}ln\left(1-x\right)\mathrm{d}x={\int }_0^1{\displaystyle \frac{ln\left(1-x\right)}{x}}{x}^{{\displaystyle \frac{n}{q}}}\mathrm{d}x,$$

(19)

Then,

$${\displaystyle \frac{\partial }{\partial q}}\left(-{\displaystyle \frac{q}{n}}{H}_{{\displaystyle \frac{n}{q}}}\right)={\int }_0^1{\displaystyle \frac{ln\left(1-x\right)}{x}}{\displaystyle \frac{\partial }{\partial q}}\left(x^{{\displaystyle \frac{n}{q}}}\right)\mathrm{d}x$$

$${\displaystyle \frac{q^2}{n^2}}{H}_{{\displaystyle \frac{n}{q}}}+{\displaystyle \frac{q}{n}}{H}_{{\displaystyle \frac{n}{q}}}^{\left(2\right)}-q\zeta \left(2\right)={\int }_0^1{\displaystyle \frac{ln\left(1-x\right)lnx}{x}}{x}^{{\displaystyle \frac{n}{q}}}\mathrm{d}x.$$

By summing over the alternating integers n for $t\in \mathbb{N}\ge 2,$

$$\begin{array}{ccc}\hfill \sum _{n\ge 1}{\displaystyle \frac{{\left(-1\right)}^n}{n^t}}\left({\displaystyle \frac{q^2}{n^2}}{H}_{{\displaystyle \frac{n}{q}}}+{\displaystyle \frac{q}{n}}{H}_{{\displaystyle \frac{n}{q}}}^{\left(2\right)}-{\displaystyle \frac{q}{n}}\zeta \left(2\right)\right)& =& {\int }_0^1{\displaystyle \frac{lnxln\left(1-x\right)}{x}}\sum _{n\ge 1}{\displaystyle \frac{{\left(-1\right)}^n{x}^{\frac{n}{q}}}{n^t}}\mathrm{d}x\hfill \\ & =& {\int }_0^1{\displaystyle \frac{lnxln\left(1-x\right)}{x}}{\mathrm{Li}}_t(-x^{{\displaystyle \frac{1}{q}}})\mathrm{d}x.\hfill \end{array}$$

By using the transformation $y^q=x$, we obtain

$$\sum _{n\ge 1}{\displaystyle \frac{{\left(-1\right)}^n}{n^t}}\left({\displaystyle \frac{q^2}{n^2}}{H}_{{\displaystyle \frac{n}{q}}}+{\displaystyle \frac{q}{n}}{H}_{{\displaystyle \frac{n}{q}}}^{\left(2\right)}-{\displaystyle \frac{q}{n}}\zeta \left(2\right)\right)=-{\int }_0^1{\displaystyle \frac{lnxln\left(1-y^q\right)}{y}}{\mathrm{Li}}_t(-y)\mathrm{d}y$$

A Taylor series expansion of the $ln\left(1-y^q\right)$ term gives us the representation

$$\sum _{n\ge 1}{\displaystyle \frac{{\left(-1\right)}^n}{n^t}}\left({\displaystyle \frac{1}{n^2}}{H}_{{\displaystyle \frac{n}{q}}}+{\displaystyle \frac{1}{qn}}{H}_{{\displaystyle \frac{n}{q}}}^{\left(2\right)}-{\displaystyle \frac{1}{qn}}\zeta \left(2\right)\right)=\sum _{n\ge 1}{\displaystyle \frac{1}{n}}{\displaystyle \underset{0}{\overset{1}{\int }}}{y}^{qn-1}lnx{\mathrm{Li}}_t(-y)\mathrm{d}y,$$

Hence, we obtain

$${\displaystyle \frac{1}{q}}\zeta \left(2\right)\eta \left(t+1\right)-{\displaystyle \frac{1}{q}}{\mathbb{S}}_{1,t+2}^{+-}({\displaystyle \frac{1}{q}})-{\displaystyle \frac{1}{q}}{\mathbb{S}}_{2,t+1}^{+-}({\displaystyle \frac{1}{q}})$$

$$={\displaystyle \frac{{\left(-1\right)}^{t}t}{2q^{t+1}}}\sum _{n\ge 1}{\displaystyle \frac{\left(\psi \left(\frac{qn}{2}+1\right)-\psi \left(\frac{qn}{2}+\frac{1}{2}\right)\right)}{n^{t+2}}}+{\displaystyle \frac{{\left(-1\right)}^t}{4q^t}}\sum _{n\ge 1}{\displaystyle \frac{\left({\psi }^{\left(\prime \right)}\left(\frac{qn}{2}+1\right)-{\psi }^{\left(\prime \right)}\left(\frac{qn}{2}+\frac{1}{2}\right)\right)}{n^{t+1}}}$$

$$+{\left(-1\right)}^{t+1}{\displaystyle \sum _{j=1}^t}{\displaystyle \frac{{\left(-1\right)}^j\left(t+1-j\right)\zeta \left(t+3-j\right)\eta \left(j\right)}{q^{t+2-j}}}.$$

From the polygamma harmonic number identity (see [2])

$${\displaystyle \frac{{\left(-1\right)}^m}{m!}}{\psi }^{\left(m\right)}\left(\alpha +1\right)=H_{\alpha }^{\left(m+1\right)}-\zeta \left(m+1\right)$$

we obtain

$${\displaystyle \frac{{\left(-1\right)}^{t}t}{q^{t+1}}}\left({\mathbb{S}}_{1,t+2}^{++}(q)-{\mathbb{S}}_{1,t+2}^{++}({\displaystyle \frac{q}{2}})\right)+{\mathbb{S}}_{1,t+2}^{+-}({\displaystyle \frac{1}{q}})+{\displaystyle \frac{{\left(-1\right)}^t}{q^t}}\left({\mathbb{S}}_{2,t+1}^{++}(q)-{\displaystyle \frac{1}{2}}{\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{q}{2}})\right)+{\displaystyle \frac{1}{q}}{\mathbb{S}}_{2,t+1}^{+-}({\displaystyle \frac{1}{q}})$$

$$\begin{array}{ccc}& =& {\displaystyle \frac{1}{q}}\zeta \left(2\right)\eta \left(t+1\right)+{\displaystyle \frac{{\left(-1\right)}^{t}t}{q^{t+1}}}\zeta \left(t+2\right)ln2\hfill \\ & & +{\displaystyle \frac{{\left(-1\right)}^t}{q^t}}\eta \left(2\right)\zeta \left(t+1\right)+{\left(-1\right)}^t{\displaystyle \sum _{j=1}^t}{\displaystyle \frac{{\left(-1\right)}^j}{q^{t+2-j}}}\left(t+1-j\right)\eta \left(j\right)\zeta (t+3-j).\hfill \end{array}$$

In the finite sum, we isolate the $j=1$ term, where $\eta \left(1\right)=ln2,$ and this completes the proof of Theorem 2. □

As a direct consequence of Theorem 2, we can highlight some special cases, as in the following Corollary.

Corollary 2.

From Theorem 2, we can highlight some special cases. For the case of $q=1,$

$${\left(-1\right)}^{t}t\left({\mathbb{S}}_{1,t+2}^{++}-{\mathbb{S}}_{1,t+2}^{++}({\displaystyle \frac{1}{2}})\right)+{\mathbb{S}}_{1,t+2}^{+-}+{\left(-1\right)}^t\left({\mathbb{S}}_{2,t+1}^{++}-{\displaystyle \frac{1}{2}}{\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{1}{2}})\right)+{\mathbb{S}}_{2,t+1}^{+-}$$

(20)

$$=\zeta \left(2\right)\eta \left(t+1\right)+{\left(-1\right)}^t\eta \left(2\right)\zeta \left(t+1\right)+{\left(-1\right)}^t{\displaystyle \sum _{j=2}^t}{\left(-1\right)}^j\left(t+1-j\right)\eta \left(j\right)\zeta (t+3-j).$$

For the case where t is even, which implies an odd weight $\left(t+3\right),$ we can write, from (20), the explicit representation

$${\displaystyle \frac{1}{2}}{\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{1}{2}})=t\left({\mathbb{S}}_{1,t+2}^{++}-{\mathbb{S}}_{1,t+2}^{++}({\displaystyle \frac{1}{2}})\right)+{\mathbb{S}}_{1,t+2}^{+-}+{\mathbb{S}}_{2,t+1}^{+-}$$

(21)

$$-\zeta \left(2\right)\eta \left(t+1\right)-\eta \left(2\right)\zeta \left(t+1\right)-{\displaystyle \sum _{j=2}^t}{\left(-1\right)}^j\left(t+1-j\right)\eta \left(j\right)\zeta (t+3-j).$$

All terms on the right of (21) are known from (6) and (8) and where, from [17] (p. 414),

$${\mathbb{S}}_{1,t+1}^{++}({\displaystyle \frac{1}{2}})={\mathbb{S}}_{1,t+1}^{+-}-{\mathbb{S}}_{1,t+1}^{++}+{\displaystyle \sum _{j=1}^{t-1}}{\displaystyle \frac{{\left(-1\right)}^{j+1}}{2^j}}\zeta (j+1)\zeta \left(t+1-j\right).$$

Specifically, for $t=4,$ we have

$${\mathbb{S}}_{2,5}^{++}({\displaystyle \frac{1}{2}})={\displaystyle \frac{107}{64}}\zeta (7)+\zeta (4)\zeta (3)-{\displaystyle \frac{21}{16}}\zeta (5)\zeta (2).$$

Considering the case where t is odd, let $t=2t^{*}-1$ (and rename $t^{*}=t$), where $t\in \mathbb{N}$, therefore

$$(1-2t)\left({\mathbb{S}}_{1,2t+1}^{++}-{\mathbb{S}}_{1,2t+1}^{++}({\displaystyle \frac{1}{2}})\right)+{\mathbb{S}}_{1,2t+1}^{+-}-\left({\mathbb{S}}_{2,2t}^{++}-{\displaystyle \frac{1}{2}}{\mathbb{S}}_{2,2t}^{++}({\displaystyle \frac{1}{2}})\right)+{\mathbb{S}}_{2,2t}^{+-}$$

$$=\zeta \left(2\right)\eta \left(2t\right)-\eta \left(2\right)\zeta \left(2t\right)-{\displaystyle \sum _{j=2}^{2t-1}}{\left(-1\right)}^j\left(2t-j\right)\eta \left(j\right)\zeta (2t+2-j).$$

By rearranging, we obtain

$${\mathbb{S}}_{2,2t}^{+-}-{\mathbb{S}}_{2,2t}^{++}+{\displaystyle \frac{1}{2}}{\mathbb{S}}_{2,2t}^{++}({\displaystyle \frac{1}{2}})+(2t-1){\mathbb{S}}_{1,2t+1}^{++}({\displaystyle \frac{1}{2}})$$

(22)

$$=(2t-2){\mathbb{S}}_{1,2t+1}^{++}+\zeta \left(2\right)\eta \left(2t\right)-\eta \left(2\right)\zeta \left(2t\right)-{\displaystyle \sum _{j=2}^{2t-1}}{\left(-1\right)}^j\left(2t-j\right)\eta \left(j\right)\zeta (2t+2-j).$$

For the specific case $t=6,$

$$\begin{array}{ccc}& & {\mathbb{S}}_{2,12}^{+-}-{\mathbb{S}}_{2,12}^{++}+{\displaystyle \frac{1}{2}}{\mathbb{S}}_{2,12}^{++}({\displaystyle \frac{1}{2}})+11{\mathbb{S}}_{1,13}^{++}({\displaystyle \frac{1}{2}})\hfill \\ & =& {\displaystyle \frac{115581}{8192}}\zeta (14)-{\displaystyle \frac{37}{64}}\zeta {(7)}^2-{\displaystyle \frac{371}{256}}\zeta (5)\zeta (9)-{\displaystyle \frac{3329}{1024}}\zeta (11)\zeta (3).\hfill \end{array}$$

3. Illustrative Examples of Euler Harmonic Sums

The next example investigates the case $q=2$ from Theorem 2 and the case of $q=4$ from Theorem 1.

Example 1.

Consider Theorem 2 and let $q=2$ with $t=2t^{*}-1$ (and rename $t^{*}=t$), where $t\in \mathbb{N}$. This allows the investigation of Euler harmonic sums with even weight. From (18),

$$\left(1-2t\right)2^{-2t}\left({\mathbb{S}}_{1,2t+1}^{++}(2)-{\mathbb{S}}_{1,2t+1}^{++}\right)+{\mathbb{S}}_{1,2t+1}^{+-}({\displaystyle \frac{1}{2}})-2^{1-2t}\left({\mathbb{S}}_{2,2t}^{++}(2)-{\displaystyle \frac{1}{2}}{\mathbb{S}}_{2,2t}^{++}\right)+{\displaystyle \frac{1}{2}}{\mathbb{S}}_{2,2t}^{+-}({\displaystyle \frac{1}{2}})$$

$$={\displaystyle \frac{1}{2}}\zeta \left(2\right)\eta \left(2t\right)-2^{1-2t}\eta \left(2\right)\zeta \left(2t\right)-{\displaystyle \sum _{j=2}^{2t-1}}{\displaystyle \frac{{\left(-1\right)}^j}{2^{2t+1-j}}}\left(2t-j\right)\eta \left(j\right)\zeta (2t+2-j).$$

(23)

By using the decomposition rule (14) and from [17] (p. 414), the half-argument identity, for $p\in \mathbb{N}\cup \left\{0\right\},t\in \mathbb{N}$

$${\displaystyle \frac{1}{2^p}}{\mathbb{S}}_{p+1,t}^{+-}({\displaystyle \frac{1}{2}})={\mathbb{S}}_{p+1,t}^{+-}-{\mathbb{S}}_{t,,p+1}^{+-}+\eta \left(p+t+1\right)-\eta \left(p+1\right)\eta \left(t\right)$$

We can simplify (23) so that

$$\left(2^{-2t}-1\right){\mathbb{S}}_{2,2t}^{++}+2{\mathbb{S}}_{2,2t}^{+-}-{\mathbb{S}}_{2t,2}^{+-}+2t{\mathbb{S}}_{1,2t+1}^{+-}={\displaystyle \frac{1}{2}}\zeta \left(2\right)\eta \left(2t\right)-2^{1-2t}\eta \left(2\right)\zeta \left(2t\right)$$

(24)

$$+{\mathbb{S}}_{2t+1,1}^{+-}+\left(2t-1\right)\left(1-2^{-2t}\right){\mathbb{S}}_{1,2t+1}^{++}-{\displaystyle \sum _{j=2}^{2t-1}}{\displaystyle \frac{{\left(-1\right)}^j}{2^{2t+1-j}}}\left(2t-j\right)\eta \left(j\right)\zeta (2t+2-j).$$

The Euler sum ${\mathbb{S}}_{2t+1,1}^{+-}$ is determined from the Alzer–Choi identity (9) and ${\mathbb{S}}_{1,2t+1}^{++}$ is the famous Euler identity (5). The identities (16) and (24) both represent Euler sums of even weight. For $t=6,$

$$\begin{array}{ccc}& & 2{\mathbb{S}}_{2,12}^{+-}-{\displaystyle \frac{4095}{4096}}{\mathbb{S}}_{2,12}^{++}-{\mathbb{S}}_{12,2}^{+-}+12{\mathbb{S}}_{1,13}^{+-}\hfill \\ & =& {\displaystyle \frac{75,989,887}{1,720,320}}\zeta (14)-{\displaystyle \frac{189}{32}}\zeta {(7)}^2-{\displaystyle \frac{375}{32}}\zeta (5)\zeta (9)-{\displaystyle \frac{5883}{512}}\zeta (11)\zeta (3).\hfill \end{array}$$

(25)

Example 2.

From Theorem 1, consider the case $q=4,$ such that

$${\displaystyle \frac{{\left(-1\right)}^t}{4^t}}\left(1+t\right){\mathbb{S}}_{1,t+2}^{++}(4)+{\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{1}{4}})$$

(26)

$$={\displaystyle \frac{{\left(-1\right)}^{t+1}}{4^{t-1}}}{\mathbb{S}}_{2,t+1}^{++}(4)+{\displaystyle \frac{1}{4}}\zeta (3)\zeta (t)+{\left(-1\right)}^{t}t{4}^{1-t}\zeta (2)\zeta (t+1)$$

$$+{\displaystyle \sum _{j=3}^t}{\left(-1\right)}^j\left({\displaystyle \frac{{\left(-1\right)}^t}{4^{t+1-j}}}\left(t+1-j\right)\zeta (j)\zeta \left(t+3-j\right)+{\displaystyle \frac{1}{4^{j-1}}}\zeta (j+1)\zeta \left(t+2-j\right)\right),$$

The Euler sum ${\mathbb{S}}_{2,t+1}^{++}(4)$ can be evaluated utilizing the method described in ref. [18], such that, for the case where t is even,

$${\mathbb{S}}_{2,t+1}^{++}(4)={\displaystyle \frac{1}{8}}\left(3\times 2^{2t+2}-3\times 2^{t+1}+3\times 2^t-1\right)\zeta (2)\zeta (t+1)$$

$$+2^{2t-1}\pi \left(t+1\right)\beta \left(t+2\right)+\left({\displaystyle \frac{1}{32}}-2^{2t+1}\left(t+2\right)-{\displaystyle \frac{\left(t+1\right)\left(t+2\right)}{64}}\right)\zeta (t+3)$$

$$+{\displaystyle \sum _{j=1}^{\left[{\displaystyle \frac{t+1}{2}}\right]}}{4}^{t+1-2j}\left(t+2-2j\right)\zeta (2j)\zeta \left(t+3-2j\right),$$

(27)

Here, the Dirichlet beta function $\beta (z)$ (see Kölbig [19]) is defined by

$$\beta (z)={\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{{(-1)}^{k+1}}{{(2k-1)}^z}}(\Re (z)>0).$$

By substituting (27) and simplifying the expression (26), we conclude that the symmetrical reciprocal identity

$${\displaystyle \frac{1}{4^t}}\left(1+t\right){\mathbb{S}}_{1,t+2}^{++}(4)+{\mathbb{S}}_{2,t+1}^{++}({\displaystyle \frac{1}{4}})={\displaystyle \frac{1}{4}}\zeta (3)\zeta (t)-2\pi \left(t+1\right)\beta \left(t+2\right)$$

$$+{\displaystyle \frac{1}{2^{2t-5}}}\left(1-3\times 2^{2t+2}+3\times 2^{t+1}-3\times 2^t+8t\right)\zeta (2)\zeta (t+1)$$

$$+{\displaystyle \frac{1}{2^{2t-2}}}\left(2^{2t+1}\left(t+2\right)+{\displaystyle \frac{\left(t+1\right)\left(t+2\right)}{64}}-{\displaystyle \frac{1}{32}}\right)\zeta (t+3)$$

$$-{\displaystyle \sum _{j=1}^{\left[{\displaystyle \frac{t+1}{2}}\right]}}{4}^{2-2j}\left(t+2-2j\right)\zeta (2j)\zeta \left(t+3-2j\right)$$

$$+{\displaystyle \sum _{j=3}^t}{\left(-1\right)}^j\left({\displaystyle \frac{1}{4^{t+1-j}}}\left(t+1-j\right)\zeta (j)\zeta \left(t+3-j\right)+{\displaystyle \frac{1}{4^{j-1}}}\zeta (j+1)\zeta \left(t+2-j\right)\right).$$

For $t=4$, we obtain

$${\displaystyle \frac{5}{4^4}}{\mathbb{S}}_{1,6}^{++}(4)+{\mathbb{S}}_{2,5}^{++}({\displaystyle \frac{1}{4}})={\displaystyle \frac{49,159}{1024}}\zeta (7)-10\pi \beta \left(6\right)-{\displaystyle \frac{5031}{512}}\zeta (2)\zeta (5)+{\displaystyle \frac{3}{16}}\zeta (4)\zeta (3).$$

4. Concluding Remarks

The scope of this study has been demonstrated in the development of Theorems 1 and 2, together with their respective corollaries. The examples have clearly demonstrated the applicability of the theorems and given new results for the linear arrangement of Euler sum identities with even weight. In particular, we have obtained closed-form identities of linear arrangements of Euler harmonic number sums of the type (4), (11), and (18) with arbitrary argument and with weight $t+3$ for $t\in \mathbb{N}.$ The linear arrangement of Euler harmonic number sums with even weight, expressed in closed form, have been given in (15), (17), (22), (24), and (25). It is expected that further research will be undertaken in the evaluation of Euler linear harmonic sums identities with multiple arguments.