Multiple Argument Euler Sum Identities

Abstract

We develop a number of new identities for linear Euler harmonic number sums with multiple argument. Some new examples of even weight linear Euler harmonic sum identities are given.

1. Introduction and Background

Euler, a notable mathematician of the seventeenth century, published many brilliant results in various branches of mathematics. In his works on series, after many illustrious scientists of his time gave various approximations to the Basel problem, Euler was able to show that

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

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, [1], 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

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

and the Dirichlet eta function, $\eta \left(t\right)$, is defined by

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

where $\Re \left(t\right)$ defines the real part of $t.$ The harmonic numbers $H_n^{\left(p\right)}$ of integer order p are defined as

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

where $\mathbb{C}$ and $\mathbb{N}$ are the sets of complex numbers and positive integers, respectively.

The harmonic numbers of order one $H_q,$ are given by the following, for $q\in {\mathbb{R}}^{+}$

$$\begin{array}{cc}{H}_q& =\gamma +\psi \left(q+1\right)\hfill \\ & =\gamma +\psi \left(q\right)+{\displaystyle \frac{1}{q}},H_0:=0.\hfill \end{array}$$

The term $\gamma$ represents the familiar Euler–Mascheroni constant (see, e.g., [1], Section 1.2), and $\psi \left(t\right)$ denotes the digamma (or psi) function defined by

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

where $\mathrm{\Gamma }\left(t\right)$ is the familiar Gamma function (see, e.g., [1], Section 1.1). Following the notation of Flajolet and Salvy [2], we define classes of linear Euler harmonic sums of the type

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

(1)

for $q\in {\mathbb{R}}^{+}\setminus \left\{0\right\}$ and with the integer $p+t$ designated as the weight. We define

$${\mathbb{S}}_{p,t}^{++}\left(1\right):={\mathbb{S}}_{p,t}^{++},{\mathbb{S}}_{p,t}^{+-}\left(1\right):={\mathbb{S}}_{p,t}^{+-}.$$

In this paper, we shall obtain linear combinations of (1) in closed form in terms of special functions, such as

$$\begin{array}{c}\hspace{1em}{\mathbb{S}}_{1,t+1}^{++}({\displaystyle \frac{1}{q}})+{\mathbb{S}}_{1,t+1}^{+-}({\displaystyle \frac{1}{q}})+{\displaystyle \frac{2{\left(-1\right)}^t}{q^t}}{\mathbb{S}}_{1,t+1}^{++}\left(q\right)-{\displaystyle \frac{{\left(-1\right)}^t}{q^t}}{\mathbb{S}}_{1,t+1}^{++}({\displaystyle \frac{q}{2}})\hfill \\ =2\sum _{j=1}^{t-1}{\displaystyle \frac{{\left(-1\right)}^{j+1}}{q^j}}\zeta \left(j+1\right)\lambda \left(t+1-j\right),\forall t\in \mathbb{N},\hfill \end{array}$$

(2)

where the Dirichlet lambda function $\lambda \left(s\right)$ is defined as the term-wise arithmetic mean of the Dirichlet eta function and the Riemann zeta function:

$$\begin{array}{cc}\lambda \left(m\right)& ={\displaystyle \frac{\eta \left(m\right)+\zeta \left(m\right)}{2}}=\underset{n\to \infty }{lim}{O}_n^{\left(m\right)}\hfill \\ & =\underset{n\to \infty }{lim}\left(\sum _{j=1}^n{\displaystyle \frac{1}{{(2j-1)}^m}}\right)\hfill \\ & =\sum _{j=1}^{\infty }{\displaystyle \frac{1}{{(2j-1)}^m}},(\Re \left(m\right)>1).\hfill \end{array}$$

For the case where $q=1$, we recall the following known results

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

(3)

due to Euler [3]. For the alternating case, for odd weight $1+t$, Sitaramachandrarao [4] published

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

(4)

and recently Alzer and Choi [5] obtained the following nice result, for $p\in \mathbb{N}\setminus \left\{1\right\}$

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

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

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

(5)

and

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

There are many interesting and significant results associated with Euler harmonic sum identities, some of which may be seen in the works of [6,7,8,9,10,11,12,13,14]. Linear Euler harmonic sums play an important role in the evaluation of integral equations in many areas of science research such as combinatorics and statistical plasma physics, especially in the context of the Sommerfeld temperature expansion of electronic entropy; see [15,16]. The majority of the published works dealing with Euler harmonic sums of the type (6) deal with the case where $q=1.$

2. Main Results

The following theorems are the main results expressing the linear Euler harmonic sums ${\mathbb{S}}_{p+1,t}^{++}\left(q\right)$ and ${\mathbb{S}}_{p+1,t}^{+-}\left(q\right)$ in terms of special functions.

Theorem 1.

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

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

(6)

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

Proof. 

Let us consider the standard integral representation for the harmonic number

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

(7)

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

$$-q\sum _{n\ge 1}{\displaystyle \frac{H_{\frac{n}{q}}}{n^{t+1}}}={\int }_0^1{\displaystyle \frac{ln\left(1-x\right)}{x}}\sum _{n\ge 1}{\displaystyle \frac{x^{\frac{n}{q}}}{n^t}}\mathrm{d}x.$$

and by (5), we obtain

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

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

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

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

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

Using the definition (5) of the Polylog function, we have that

$$\begin{array}{cc}\sum _{n\ge 1}{\displaystyle \frac{H_{\frac{n}{q}}}{n^{t+1}}}& =\sum _{n\ge 1}{\displaystyle \frac{1}{n}}\sum _{k\ge 1}{\displaystyle \frac{1}{k^t}}\underset{0}{\overset{1}{\int }}{y}^{qn+k-1}\mathrm{d}y\hfill \\ & =\sum _{n\ge 1}{\displaystyle \frac{1}{n}}\sum _{k\ge 1}{\displaystyle \frac{1}{k^t\left(qn+k\right)}}\hfill \\ & =\sum _{n\ge 1}{\displaystyle \frac{1}{n}}\sum _{k\ge 1}\left({\displaystyle \frac{{\left(-1\right)}^{t+1}}{{\left(qn\right)}^{t-1}k\left(qn+k\right)}}+\sum _{j=1}^{t-1}{\displaystyle \frac{{\left(-1\right)}^{j+1}}{{\left(nq\right)}^j{k}^{t+1-j}}}\right),\hfill \end{array}$$

which is simplified to

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

and the proof of the theorem is complete. Alternatively, the proof may also be achieved by considering the known identity

$$H_q=\sum _{\rho \ge 1}\left({\displaystyle \frac{1}{\rho }}-{\displaystyle \frac{1}{\rho +q}}\right).$$

Then,

$$\begin{array}{cc}\sum _{n\ge 1}{\displaystyle \frac{H_{\frac{n}{q}}}{n^{t+1}}}& =\sum _{n\ge 1}{\displaystyle \frac{1}{n^t}}\sum _{\rho \ge 1}\left({\displaystyle \frac{1}{\rho \left(n+q\rho \right)}}\right)=\sum _{\rho \ge 1}{\displaystyle \frac{1}{\rho }}\sum _{n\ge 1}{\displaystyle \frac{1}{n^t}}{\displaystyle \frac{1}{\left(n+q\rho \right)}}\hfill \\ & =\sum _{n\ge 1}{\displaystyle \frac{1}{n}}\underset{0}{\overset{1}{\int }}{y}^{qn-1}{\mathrm{Li}}_t\left(y\right)\mathrm{d}y.\hfill \end{array}$$

□

Corollary 1.

From Theorem 1, we can highlight some special cases. Let us consider the case of even weight. Let $t=2t$; hence,

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

(8)

Rearranging (8), we can isolate the interesting identity, for $t\in \mathbb{N}$ and $q\in {\mathbb{R}}^{+},$

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

For the case where $q=1$, we recover the result obtained by Georghiou and Philippou [17], namely,

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

(9)

From (6), for the case where $q=2$ and $t\in \mathbb{N}$, we obtain

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

(10)

Applying the decomposition relation

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

(11)

we can rewrite (10) as

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

where ${\mathbb{S}}_{1,t+1}^{++}$ is given by (3).

For even weight $\left(2t+2\right)$, we obtain the identity

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

For $t=1,$

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

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

$${\mathbb{S}}_{1,3}^{++}({\displaystyle \frac{1}{2}})={\mathbb{S}}_{1,3}^{+-}={\displaystyle \frac{11}{4}}\zeta \left(4\right)-{\displaystyle \frac{7}{4}}\zeta \left(3\right)ln2+{\displaystyle \frac{1}{2}}\zeta \left(2\right){ln}^{2}2-{\displaystyle \frac{1}{12}}{ln}^{4}2-2{\mathrm{Li}}_4({\displaystyle \frac{1}{2}}).$$

For $t=6,$

$$\begin{array}{cc}{\mathbb{S}}_{1,13}^{++}({\displaystyle \frac{1}{2}})-{\mathbb{S}}_{1,13}^{+-}& =-{\mathbb{S}}_{1,2t+1}^{++}+\sum _{j=1}^{2t-1}{\displaystyle \frac{{\left(-1\right)}^{j+1}}{2^j}}\zeta (j+1)\zeta \left(2t+1-j\right)\hfill \\ & ={\displaystyle \frac{767}{1024}}\zeta \left(3\right)\zeta \left(11\right)+{\displaystyle \frac{239}{256}}\zeta \left(5\right)\zeta \left(9\right)+{\displaystyle \frac{31}{64}}{\zeta }^2\left(7\right)-{\displaystyle \frac{22521}{8192}}\zeta \left(14\right).\hfill \end{array}$$

In the case of t being odd, let $t=2t-1$, so that

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

(12)

and if we now consider the case where $q=2$ and use the decomposition identity (11), we have

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

(13)

where ${\mathbb{S}}_{1,2t}^{++}$ is given by (3) and ${\mathbb{S}}_{1,2t}^{+-}$ is the Sitaramachandrarao identity (4). If we rewrite (12) as

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

and choose $q=1,$ we obtain

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

where ${\mathbb{S}}_{1,2t}^{++}$ is the Euler identity and ${\mathbb{S}}_{1,2t}^{++}({\displaystyle \frac{1}{2}})$ is given by (13), which evidently implies that

$$\sum _{j=1}^{2t-2}{\left(-1\right)}^{j+1}\zeta (j+1)\zeta \left(2t-j\right)=0.$$

The next theorem deals with the Euler sum ${\mathbb{S}}_{1,t+1}^{+-}({\displaystyle \frac{1}{q}}).$

Theorem 2.

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

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

(14)

where ${\mathbb{S}}_{1,t+1}^{+-}\left(q\right)$ is given by (1), of weight $t+2$, and $\eta \left(t\right)$ is the classical Riemann eta function.

Proof. 

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

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

Summing over the alternating integers n,

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

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

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

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

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

Now,

$$\begin{array}{cc}\sum _{n\ge 1}{\displaystyle \frac{H_{\frac{n}{q}}}{n^{t+1}}}& =\sum _{n\ge 1}{\displaystyle \frac{1}{n}}\sum _{k\ge 1}{\displaystyle \frac{{\left(-1\right)}^k}{k^t}}\underset{0}{\overset{1}{\int }}{y}^{qn+k-1}\mathrm{d}y\hfill \\ & =\sum _{n\ge 1}{\displaystyle \frac{1}{n}}\sum _{k\ge 1}{\displaystyle \frac{{\left(-1\right)}^k}{k^t\left(qn+k\right)}}\hfill \\ & =\sum _{n\ge 1}{\displaystyle \frac{1}{n}}\sum _{k\ge 1}{\left(-1\right)}^k\left({\displaystyle \frac{{\left(-1\right)}^{t+1}}{{\left(qn\right)}^{t-1}k\left(qn+k\right)}}+\sum _{j=1}^{t-1}{\displaystyle \frac{{\left(-1\right)}^{j+1}}{{\left(nq\right)}^j{k}^{t+1-j}}}\right)\hfill \end{array}$$

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

From the multiple argument identity of harmonic numbers

$$H_{{\displaystyle \frac{nq}{2}}-{\displaystyle \frac{1}{2}}}=2H_{nq}-2ln2-H_{{\displaystyle \frac{nq}{2}}},$$

we obtain

$$\sum _{n\ge 1}{\displaystyle \frac{{\left(-1\right)}^{n+1}{H}_{\frac{n}{q}}}{n^{t+1}}}={\displaystyle \frac{{\left(-1\right)}^t}{q^t}}\sum _{n\ge 1}\left(H_{{\displaystyle \frac{nq}{2}}}+ln2-H_{nq}\right)+\sum _{j=1}^t{\displaystyle \frac{{\left(-1\right)}^{j+1}\zeta \left(j+1\right)\eta \left(t+1-j\right)}{q^j}}.$$

Isolating the $j=t$ term and noting that $\eta \left(1\right)=ln2,$ we have

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

and this completes the proof of Theorem 2. □

The next corollary deals with some special cases of Theorem 2

Corollary 2.

From Theorem 2, we can highlight some special cases. Adding cases (6) and (14), we may write

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

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

which confirms result (2). If $q=1,$ (14) yields

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

If $q=2,$ (14) yields

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

and utilizing the decomposition rule (11), we obtain, for $t\in \mathbb{N},$

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

(15)

For odd weight $2t+1$, we may write

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

and employing the known identities (3) and (4), after some algebraic simplification, we have

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

From the decomposition Formula (11), we obtain a similar identity for ${\mathbb{S}}_{1,2t}^{++}\left({\displaystyle \frac{1}{2}}\right).$

For even weight $2t+2$, we may write

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

(16)

The individual sums on the left hand side of (16) are unknown in closed form, except for the case where $t=1,$ in which case

$$\begin{array}{cc}{\mathbb{S}}_{1,3}^{+-}({\displaystyle \frac{1}{2}})& ={\mathbb{S}}_{1,3}^{+-}-{\displaystyle \frac{1}{8}}{\zeta }^2\left(2\right)\hfill \\ & ={\displaystyle \frac{39}{16}}\zeta \left(4\right)-{\displaystyle \frac{7}{4}}\zeta \left(3\right)ln2+{\displaystyle \frac{1}{2}}\zeta \left(2\right){ln}^{2}2-{\displaystyle \frac{1}{12}}{ln}^{4}2-2{\mathrm{Li}}_4({\displaystyle \frac{1}{2}}),\hfill \end{array}$$

where ${\mathrm{Li}}_4(·)$ is the polylogarithm function defined by (5).

For $t=6,$

$${\mathbb{S}}_{1,13}^{+-}({\displaystyle \frac{1}{2}})-{\mathbb{S}}_{1,13}^{+-}={\displaystyle \frac{3069}{4096}}\zeta \left(3\right)\zeta \left(11\right)+{\displaystyle \frac{3825}{4096}}\zeta \left(5\right)\zeta \left(9\right)+{\displaystyle \frac{3969}{8192}}{\zeta }^2\left(7\right)-{\displaystyle \frac{45057}{8192}}\zeta \left(14\right).$$

3. Concluding Remarks

We have obtained closed-form identities of linear combinations of Euler harmonic number sums of the type (2), (6), and (14) with arbitrary argument and with weight $t+2$ for $t\in \mathbb{N}.$ Some results previously published by Georghiou and Philippou (see (9)) have been recovered. Some new examples of even-weight linear Euler harmonic sum identities are given.