## Background and preliminaries

In the spirit of Euler, we shall investigate the summation of some variant Euler sums. In common terminology, let, as usual,

${H}_{n}=\gamma +\psi \left(n+1\right)=\sum _{r=1}^{n}\frac{1}{r}=\underset{0}{\overset{\infty }{\int }}\frac{1-{t}^{n}}{1-t}\mathrm{dt}$
(1)

be the n th harmonic number, γ denotes the Euler-Mascheroni constant, $\psi \left(z\right):=dlog\Gamma \left(z\right)/\mathrm{dz}$ is the digamma function and $\Gamma \left(z\right)$ is the well-known gamma function. Let also, $ℝ,\phantom{\rule{1em}{0ex}}ℂ$ and $ℕ$ denote, respectively, the sets of real, complex and natural numbers. A generalized binomial coefficient $\left(\genfrac{}{}{0}{}{w}{z}\right)$ may be defined by

$\left(\genfrac{}{}{0}{}{w}{z}\right):=\frac{\Gamma \left(w+1\right)}{\Gamma \left(z+1\right)\Gamma \left(w-z+1\right)}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(w,\phantom{\rule{1em}{0ex}}z\in ℂ\right)$
(2)

and in the special case when $z=n,\phantom{\rule{1em}{0ex}}n\in ℕ,$ we have

$\left(\genfrac{}{}{0}{}{w}{n}\right):=\frac{w\left(w-1\right)\mathrm{...}\left(w-n+1\right)}{n!}=\frac{{\left(-1\right)}^{n}{\left(-w\right)}_{n}}{n!},$
(3)

where

$\phantom{\rule{-14.0pt}{0ex}}{\left(w\right)}_{\lambda }:=\phantom{\rule{0.3em}{0ex}}\frac{\Gamma \left(w\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\lambda \right)}{\Gamma \left(w\right)}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\left\{\phantom{\rule{0.3em}{0ex}}\begin{array}{c}1\left(\lambda \phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}0;\phantom{\rule{1em}{0ex}}w\in ℂ\setminus \left\{0\right\}\right)\\ w\left(w\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}1\right)& \dots \left(\phantom{\rule{0.3em}{0ex}}w\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\lambda \phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}1\phantom{\rule{0.3em}{0ex}}\right)\phantom{\rule{0.3em}{0ex}}\left(\phantom{\rule{0.3em}{0ex}}w\in \phantom{\rule{0.3em}{0ex}}ℂ,\phantom{\rule{1em}{0ex}}\lambda \phantom{\rule{0.3em}{0ex}}\in \phantom{\rule{0.3em}{0ex}}ℕ\right)\end{array}\right\$
(4)

with ${\left(0\right)}_{0}:=1$ is known as the Pochhammer symbol. Some well-known Euler sums are (see, e.g., [1])

$\sum _{n=1}^{\infty }\frac{{H}_{n}}{{n}^{2}}=2\zeta \left(3\right),\phantom{\rule{1em}{0ex}}\sum _{n=1}^{\infty }{\left(\frac{{H}_{n}}{n}\right)}^{2}=\frac{17}{4}\zeta \left(4\right);$
(5)

recently, Chen [2] obtained

$\sum _{n=1}^{\infty }\frac{{H}_{2n}}{{\left(2n+1\right)}^{2}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\frac{7}{16}\zeta \left(3\right),\phantom{\rule{1em}{0ex}}\sum _{n=1}^{\infty }\frac{{H}_{2n}-\frac{1}{2}{H}_{n}}{{4}^{n}n}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\frac{1}{4}{\left(ln3\right)}^{2}.$
(6)

In [3], we have, for k≥1,

$\begin{array}{cc}\sum _{n-1}^{\infty }\frac{{H}_{n}}{n\left(\genfrac{}{}{0}{}{n+k}{k}\right)}& =2\zeta \left(2\right)+\sum _{r=1}^{k}{\left(-1\right)}^{r+1}\left(\genfrac{}{}{0}{}{k}{r}\right)\\ \phantom{\rule{1em}{0ex}}×\left({H}_{r-1}^{2}+{H}_{r-1}^{\left(2\right)}\right)\end{array}$
(7)

and in [4],

$\begin{array}{cc}\sum _{n-1}^{\infty }\frac{\phantom{\rule{1em}{0ex}}{H}_{n}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+k}{k}\right)}& ={H}_{p}^{\left(2\right)}+{H}_{p}^{2}+\sum _{r=1}^{k}{\left(-1\right)}^{r}\left(\genfrac{}{}{0}{}{k}{r}\right)\\ \phantom{\rule{1em}{0ex}}×\left(\genfrac{}{}{0}{}{{H}_{p}{H}_{p+r}-{H}_{p}{H}_{r-1}}{-\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\left(\frac{{H}_{m}-{H}_{m+p}}{m}\right)}\right),\end{array}$
(8)

where ${H}_{n}^{\left(r\right)}$ denotes the generalized nth harmonic number in power r defined by

${H}_{n}^{\left(r\right)}:=\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{n}\frac{1}{{m}^{r}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(n,r\in ℕ\right).$
(9)

We study, in this paper, $\sum _{n=1}^{\infty }\frac{{H}_{2n}}{n\left(\genfrac{}{}{0}{}{n+k}{k}\right)}$ and its finite counterpart. Analogous results of Euler type for infinite series have been developed by many authors, see for example [5, 6] and references therein. Many finite versions of harmonic number sum identities also exist in the literature, for example in [7], we have

$\sum _{n=0}^{p}{\left(-1\right)}^{n}\left(\genfrac{}{}{0}{}{p}{n}\right){H}_{n+b}^{2}=\frac{2{H}_{p-1}-{H}_{p+b}-{H}_{b}}{p\left(\genfrac{}{}{0}{}{p+b}{b}\right)},$
(10)

and in [8],

$\phantom{\rule{-13.0pt}{0ex}}\sum _{n=0}^{p}{\left(\phantom{\rule{0.3em}{0ex}}-1\phantom{\rule{0.3em}{0ex}}\right)}^{n}\left(\genfrac{}{}{0}{}{p+n}{n}\right)\left(\genfrac{}{}{0}{}{p}{n}\right)n{H}_{n}={\left(-1\right)}^{p}p\left(p+1\right)\left(2{H}_{p}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}1\right)\phantom{\rule{0.3em}{0ex}}.$
(11)

Also, from the study of Prodinger [9],

$\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\left(\genfrac{}{}{0}{}{n+k}{k}\right){\left(-1\right)}^{n-k}{H}_{k}=\sum _{m=1}^{n}\frac{2}{m}=2{H}_{n}.$
(12)

Further work in the summation of harmonic numbers and binomial coefficients has also been done by Sofo [10]. The works of [1117] and references therein also investigate various representations of binomial sums and zeta functions in a simpler form by the use of the beta function and by means of certain summation theorems for hypergeometric series.

### Lemma 1

Let n and r be positive integers. Then we have

$\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{n}\frac{\phantom{\rule{1em}{0ex}}1\phantom{\rule{1em}{0ex}}}{2r-1}=\frac{1}{2}{H}_{n-\frac{1}{2}}+ln2$
(13)
$={H}_{2n}-\frac{1}{2}{H}_{n};$
(14)
$\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{\phantom{\rule{1em}{0ex}}{H}_{m}\phantom{\rule{1em}{0ex}}}{2m+1}=\frac{1}{2}{H}_{r-1}{H}_{r-\frac{1}{2}}-\frac{1}{2}\sum _{s\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{{H}_{s-\frac{1}{2}}}{s};$
(15)
$\begin{array}{l}\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r}\left(\frac{\phantom{\rule{1em}{0ex}}{H}_{m}\phantom{\rule{1em}{0ex}}}{2m+1}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\frac{\phantom{\rule{1em}{0ex}}{H}_{2m}\phantom{\rule{1em}{0ex}}}{m}\right)\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{H}_{r}{H}_{2r+1}+\frac{1}{4}\left({H}_{r}^{\left(2\right)}-{H}_{r}^{2}\right)\phantom{\rule{0.3em}{0ex}}.\end{array}$
(16)

### Proof

From the definition of harmonic numbers and the digamma function,

${H}_{n-\frac{1}{2}}=\gamma +\psi \left(n+\frac{1}{2}\right)=\gamma +2\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{n}\frac{\phantom{\rule{1em}{0ex}}1\phantom{\rule{1em}{0ex}}}{2r-1}-\gamma -2ln2$
(17)

and Equation 3 follows. From the double argument identity of the digamma function

$\begin{array}{ll}\psi \left(2n\right)& =\frac{1}{2}\psi \left(n\right)+\frac{1}{2}\psi \left(n+\frac{1}{2}\right)+ln2\phantom{\rule{2em}{0ex}}\\ {H}_{2n-1}-\gamma & =\frac{1}{2}\left({H}_{n-1}-\gamma \right)+\frac{1}{2}\left({H}_{n-\frac{1}{2}}-\gamma \right)+ln2,\phantom{\rule{2em}{0ex}}\end{array}$
(18)

using Equation 3 and rearranging, we obtain Equation 4. For Equation 5, we first note that for an arbitrary sequence $\left\{{X}_{k,l\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}}\right\}$, the following identity holds:

$\sum _{k\phantom{\rule{0.3em}{0ex}}=1}^{n}\sum _{l\phantom{\rule{0.3em}{0ex}}=1}^{k}{X}_{k,l}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}=\sum _{l\phantom{\rule{0.3em}{0ex}}=1}^{n}\sum _{k\phantom{\rule{0.3em}{0ex}}=l}^{n}{X}_{k,l}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}};$
(19)

hence,

$\begin{array}{ll}\phantom{\rule{-13.0pt}{0ex}}\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{{H}_{m}}{2m+1}& =\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\sum _{s\phantom{\rule{0.3em}{0ex}}=1}^{m}\frac{1}{s\left(2m+1\right)}=\sum _{s\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\sum _{m=s\phantom{\rule{0.3em}{0ex}}}^{r-1}\frac{1}{s\left(2m+1\right)}\phantom{\rule{2em}{0ex}}\\ =\sum _{s\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{1}{2s}\left({H}_{r-\frac{1}{2}}-{H}_{s-\frac{1}{2}}\right)=\frac{1}{2}{H}_{r-1}{H}_{r-\frac{1}{2}}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\frac{1}{2}\sum _{s\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{\phantom{\rule{1em}{0ex}}{H}_{s-\frac{1}{2}}}{s}.\phantom{\rule{2em}{0ex}}\end{array}$
(20)

The interesting identity (Equation 6) follows from Equation 5 and substituting

$\frac{1}{2}{H}_{n-\frac{1}{2}}={H}_{2n}-\frac{1}{2}{H}_{n}-ln2$
(21)

so that

$\begin{array}{ll}\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{\phantom{\rule{1em}{0ex}}{H}_{m}\phantom{\rule{1em}{0ex}}}{2m+1}& ={H}_{r-1}\left({H}_{2r}-\frac{1}{2}{H}_{r}-ln2\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{\left({H}_{2m}-\frac{1}{2}{H}_{m}-ln2\right)}{m}\phantom{\rule{2em}{0ex}}\end{array}$
(22)
$\begin{array}{l}={H}_{r-1}\left({H}_{2r}-\frac{1}{2}{H}_{r}-ln2\right)+{H}_{r-1}ln2\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\frac{1}{2}\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{{H}_{m}}{m}-\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{{H}_{2m}}{m}\phantom{\rule{2em}{0ex}}\\ ={H}_{r-1}\left({H}_{2r}-\frac{1}{2}\left(\frac{1}{r}+{H}_{r-1}\right)\right)+\frac{1}{4}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\left({H}_{r-1}^{2}+{H}_{r-1}^{\left(2\right)}\right)-\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{{H}_{2m}}{m}\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\left(\frac{\phantom{\rule{1em}{0ex}}{H}_{m}\phantom{\rule{1em}{0ex}}}{2m+1}+\frac{{H}_{2m}}{m}\right)={H}_{r-1}{H}_{2r-1}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\frac{1}{4}\left({H}_{r-1}^{\left(2\right)}-{H}_{r-1}^{2}\right)\phantom{\rule{2em}{0ex}}\end{array}$
(23)

replacing the counter, we obtain Equation 6. □

## Main results and discussion

We now prove the two following theorems:

### Theorem 1

Let $k\in \mathrm{ℕ.}$ Then we have

$\phantom{\rule{-15.0pt}{0ex}}\begin{array}{cc}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{{H}_{2n}-\frac{1}{2}\phantom{\rule{1em}{0ex}}{H}_{n}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+k}{k}\right)}& =\sum _{r=1}^{k}{\left(-1\right)}^{r+1}\left(\genfrac{}{}{0}{}{k}{r}\right)\\ \phantom{\rule{1em}{0ex}}×\left(\phantom{\rule{0.3em}{0ex}}\begin{array}{c}{H}_{r-1}{H}_{2r-1}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\frac{1}{4}\left({H}_{r-1}^{\left(\phantom{\rule{0.3em}{0ex}}2\phantom{\rule{0.3em}{0ex}}\right)}-{H}_{r-1}^{2}\right)\\ \phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}2ln2\left({H}_{2r}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\frac{1}{2}{H}_{r}\right)\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\sum _{j=1}^{r-1}\frac{{H}_{2j}}{j}\end{array}\phantom{\rule{0.3em}{0ex}}\right)\phantom{\rule{0.3em}{0ex}}.\end{array}$
(24)

### Proof

Let ${h}_{n}={H}_{2n}-\frac{1}{2}\phantom{\rule{1em}{0ex}}{H}_{n}$ and consider the following expansion:

$\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{{h}_{n}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+k}{k}\right)}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{k!\phantom{\rule{1em}{0ex}}{h}_{n}\phantom{\rule{1em}{0ex}}}{n\prod _{r=1}^{k}\left(n+r\right)}=\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{k!\phantom{\rule{1em}{0ex}}{h}_{n}\phantom{\rule{1em}{0ex}}}{n\phantom{\rule{1em}{0ex}}{\left(n+1\right)}_{k+1}}.$
(25)

Now,

$\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{\phantom{\rule{1em}{0ex}}{h}_{n}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+k}{k}\right)}=\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{k!\phantom{\rule{1em}{0ex}}{h}_{n}\phantom{\rule{1em}{0ex}}}{n}\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}\frac{{A}_{r}\phantom{\rule{1em}{0ex}}}{n+r},$
(26)

where

$\begin{array}{l}{A}_{r}=\underset{n\to -r}{lim}\frac{n+r}{\prod _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}n+r}=\frac{{\left(-1\right)}^{r+1}\phantom{\rule{1em}{0ex}}r}{k!}\phantom{\rule{0.3em}{0ex}}\left(\genfrac{}{}{0}{}{k}{r}\right)\phantom{\rule{0.3em}{0ex}}.\end{array}$
(27)

For an arbitrary positive sequence $\left\{{X}_{k,p\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}}\right\}$, the following identity holds:

$\sum _{k\phantom{\rule{0.3em}{0ex}}=0}^{\infty }\sum _{p\phantom{\rule{0.3em}{0ex}}=0}^{n}{X}_{p,k}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}=\sum _{k\phantom{\rule{0.3em}{0ex}}=0}^{\infty }\sum _{p\phantom{\rule{0.3em}{0ex}}=0}^{\infty }{X}_{p,k+p}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}};$
(28)

hence, from Equations 4 and 9,

$\begin{array}{ll}\phantom{\rule{-14.0pt}{0ex}}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{k!\phantom{\rule{1em}{0ex}}{h}_{n}\phantom{\rule{1em}{0ex}}}{n}\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}\frac{{A}_{r}\phantom{\rule{1em}{0ex}}}{n+r}& =\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}\phantom{\rule{1em}{0ex}}r\phantom{\rule{1em}{0ex}}\left(\genfrac{}{}{0}{}{k}{r}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{1\phantom{\rule{1em}{0ex}}}{n\left(n+r\right)}\sum _{j\phantom{\rule{0.3em}{0ex}}=1}^{n}\frac{1\phantom{\rule{1em}{0ex}}}{2j-1}\phantom{\rule{2em}{0ex}}\\ =\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}\phantom{\rule{1em}{0ex}}r\phantom{\rule{1em}{0ex}}\left(\genfrac{}{}{0}{}{k}{r}\right)\sum _{j\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{1\phantom{\rule{1em}{0ex}}}{2j-1}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\sum _{n\phantom{\rule{0.3em}{0ex}}=0}^{\infty }\frac{1\phantom{\rule{1em}{0ex}}}{\left(n+j\right)\left(n+j+r\right)}\phantom{\rule{2em}{0ex}}\\ =\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}\phantom{\rule{1em}{0ex}}r\phantom{\rule{1em}{0ex}}\left(\genfrac{}{}{0}{}{k}{r}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\sum _{j\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{1\phantom{\rule{1em}{0ex}}}{2j-1}\left(\frac{\psi \left(j+r\right)\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\psi \left(r\right)}{r}\right)\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}\end{array}$
(29)

Since we notice that

$\frac{\psi \left(j+r\right)-\psi \left(r\right)}{r}=\frac{1}{j}+\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{1\phantom{\rule{1em}{0ex}}}{m+j},$
(30)

we get

$\begin{array}{ll}\phantom{\rule{-13.0pt}{0ex}}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{\phantom{\rule{1em}{0ex}}{h}_{n}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+k}{k}\right)}& =\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(\genfrac{}{}{0}{}{k}{r}\right)\sum _{j\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{1\phantom{\rule{1em}{0ex}}}{2j-1}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\left(\frac{1}{j}+\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{1\phantom{\rule{1em}{0ex}}}{m+j}\right)=\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\left(\genfrac{}{}{0}{}{k}{r}\right)\left(2ln2+\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{2ln2+{H}_{m}\phantom{\rule{1em}{0ex}}}{2m+1}\right)\phantom{\rule{2em}{0ex}}\\ =\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(\genfrac{}{}{0}{}{k}{r}\right)\left(\phantom{\rule{-52.0pt}{0ex}}2ln2+2ln2\right\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\left(-1\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}ln2\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\frac{1}{2}{H}_{r-\frac{1}{2}}\phantom{\rule{0.3em}{0ex}}\right)\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\phantom{\rule{0.3em}{0ex}}\frac{{H}_{m}\phantom{\rule{1em}{0ex}}}{2m+1}\phantom{\rule{0.3em}{0ex}})\phantom{\rule{2em}{0ex}}\\ =2\stackrel{2}{ln}\left(2\right)+\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}\left(\genfrac{}{}{0}{}{k}{r}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\left(ln2{H}_{r-\frac{1}{2}}+\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{{H}_{m}\phantom{\rule{1em}{0ex}}}{2m+1}\right).\phantom{\rule{2em}{0ex}}\end{array}$
(31)

Now,

$\begin{array}{ll}\phantom{\rule{-13.0pt}{0ex}}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{{h}_{n}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+k}{k}\right)}& =\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(\genfrac{}{}{0}{}{k}{r}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\left(\phantom{\rule{0.3em}{0ex}}\begin{array}{l}ln2{H}_{r-\frac{1}{2}}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\frac{1}{2}{H}_{r-1}{H}_{r-\frac{1}{2}}\\ -\frac{1}{2}\sum _{s\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{\phantom{\rule{1em}{0ex}}{H}_{s-\frac{1}{2}}\phantom{\rule{1em}{0ex}}}{s}\end{array}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\right)\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}2{ln}^{2}\phantom{\rule{0.3em}{0ex}}\left(2\right)\phantom{\rule{2em}{0ex}}\\ =\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(\genfrac{}{}{0}{}{k}{r}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\left(\begin{array}{l}\frac{1}{2}{H}_{r-1}{H}_{r-\frac{1}{2}}-\frac{1}{2}\sum _{s\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{{H}_{s-\frac{1}{2}}\phantom{\rule{1em}{0ex}}}{s}\\ +ln2\left(-2ln2+2\sum _{j=1}^{r}\frac{1}{2j-1}\right)\end{array}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+2\stackrel{2}{ln}\left(2\right)=\sum _{r=1}^{k}{\left(-1\right)}^{r+1}\left(\genfrac{}{}{0}{}{k}{r}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\left(\frac{1}{2}{H}_{r-1}{H}_{r-\frac{1}{2}}+\sum _{j=1}^{r}\frac{2ln2\phantom{\rule{1em}{0ex}}}{2j-1}\right\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\frac{1}{2}\sum _{j\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{{H}_{j-\frac{1}{2}}\phantom{\rule{1em}{0ex}}}{j});\phantom{\rule{2em}{0ex}}\end{array}$
(32)

substituting Equation 7 and simplifying, we have

$\phantom{\rule{-15.0pt}{0ex}}\begin{array}{c}=\sum _{r=1}^{k}{\left(-1\right)}^{r+1}\left(\genfrac{}{}{0}{}{k}{r}\right)\\ \phantom{\rule{1em}{0ex}}×\left(\phantom{\rule{0.3em}{0ex}}\begin{array}{l}{H}_{r-1}\left({H}_{2r}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\frac{1}{2}\left({H}_{r-1}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\frac{1}{r}\right)\right)+2ln2\left({H}_{2r}-\frac{1}{2}{H}_{r}\right)\\ +\frac{1}{4}\left({H}_{r-1}^{2}+{H}_{r-1}^{\left(2\right)}\right)\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\sum _{j\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{{H}_{2j}\phantom{\rule{1em}{0ex}}}{j}\end{array}\phantom{\rule{0.3em}{0ex}}\right)\phantom{\rule{0.3em}{0ex}};\end{array}$
(33)

hence, the identity (Equation 8) follows. □

### Corollary 1

From Equation 8 and using Equations 3 and 4, we obtain the results,

$\begin{array}{ll}\phantom{\rule{-14.0pt}{0ex}}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{{H}_{2n}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+k}{k}\right)}& =\sum _{r=1}^{k}{\left(-1\right)}^{r+1}\left(\genfrac{}{}{0}{}{k}{r}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\left(\phantom{\rule{0.3em}{0ex}}\begin{array}{l}{H}_{r-1}{H}_{2r-1}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\frac{1}{4}\left(5{H}_{r-1}^{\left(2\right)}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}3{H}_{r-1}^{2}\right)\\ +2ln2\left({H}_{2r}-\frac{1}{2}{H}_{r}\right)-\sum _{j=1}^{r-1}\frac{{H}_{2j}}{j}\end{array}\phantom{\rule{0.3em}{0ex}}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\frac{\zeta \left(2\right)}{2}\phantom{\rule{2em}{0ex}}\end{array}$
(34)

and

$\begin{array}{ll}\phantom{\rule{-14.0pt}{0ex}}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{{H}_{n-\frac{1}{2}}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+k}{k}\right)}& =2\sum _{r=1}^{k}{\left(-1\right)}^{r+1}\left(\genfrac{}{}{0}{}{k}{r}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\left(\phantom{\rule{0.3em}{0ex}}\begin{array}{l}{H}_{r-1}{H}_{2r-1}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\frac{1}{4}\left({H}_{r-1}^{\left(2\right)}-{H}_{r-1}^{2}\right)\\ +2ln2\left({H}_{2r}-\frac{1}{2}{H}_{r}\right)-\sum _{j=1}^{r-1}\frac{{H}_{2j}}{j}\end{array}\phantom{\rule{0.3em}{0ex}}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\frac{2ln2}{k}.\phantom{\rule{2em}{0ex}}\end{array}$
(35)

### Proof

We can use Equations 3 and 4 and also note that

$\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{ln2\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+k}{k}\right)}=\frac{ln2\phantom{\rule{1em}{0ex}}}{k}.$
(36)

From the rearrangement of $\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{{H}_{2n}-\frac{1}{2}\phantom{\rule{1em}{0ex}}{H}_{n}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+k}{k}\right)}$ and Equation 1, we can obtain Equation 11; and from the rearrangement of $\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{\frac{1}{2}{H}_{n-\frac{1}{2}}+ln2\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+k}{k}\right)}$ and Equation 13, we can obtain Equation 12. □

### Example 1

For k=3 and 5,

$\begin{array}{ll}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{{H}_{n-\frac{1}{2}}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+3}{3}\right)}& =\frac{22ln2}{15}-\frac{11}{15},\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{{H}_{2n}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+3}{3}\right)}\phantom{\rule{2em}{0ex}}\\ =\frac{\zeta \left(2\right)}{2}+\frac{16ln2}{15}-\frac{119}{120}\phantom{\rule{2em}{0ex}}\\ \sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{{H}_{n-\frac{1}{2}}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+5}{5}\right)}& =\frac{386ln2}{315}-\frac{1321}{1890},\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{{H}_{2n}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+5}{5}\right)}\phantom{\rule{2em}{0ex}}\\ =\frac{\zeta \left(2\right)}{2}+\frac{256ln2}{315}-\frac{32093}{30240}\phantom{\rule{2em}{0ex}}\\ \sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{{H}_{2n}-\frac{1}{2}{H}_{n}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+3}{3}\right)}& =\frac{16ln2}{15}-\frac{11}{30},\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{{H}_{2n}-\frac{1}{2}{H}_{n}\phantom{\rule{1em}{0ex}}}{n\left(\genfrac{}{}{0}{}{n+5}{5}\right)}\phantom{\rule{2em}{0ex}}\\ =\frac{256ln2}{315}-\frac{1321}{3780}.\phantom{\rule{2em}{0ex}}\end{array}$
(37)

Now, we consider the following finite version of Theorem 1:

### Theorem 2

Let k, $p\in \mathrm{ℕ.}$ Then we have

$\begin{array}{ll}\phantom{\rule{-15.0pt}{0ex}}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{p}\frac{{H}_{2n}-\frac{1}{2}\phantom{\rule{1em}{0ex}}{H}_{n}\phantom{\rule{1em}{0ex}}}{n\left(\begin{array}{l}n+k\\ k\end{array}\right)}& =\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{p}\frac{\frac{1}{2}{H}_{n-\frac{1}{2}}+ln2\phantom{\rule{1em}{0ex}}}{n\left(\begin{array}{l}n+k\\ k\end{array}\right)}\phantom{\rule{2em}{0ex}}\\ ={H}_{p}\phantom{\rule{0.3em}{0ex}}\left(\phantom{\rule{0.3em}{0ex}}{H}_{2p}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\frac{1}{2}{H}_{p}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}1\phantom{\rule{0.3em}{0ex}}\right)\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\sum _{r=1}^{k}{\left(-1\right)}^{r+1}\phantom{\rule{0.3em}{0ex}}\left(\phantom{\rule{0.3em}{0ex}}\begin{array}{l}k\\ r\end{array}\phantom{\rule{0.3em}{0ex}}\right)\left(\phantom{\rule{0.3em}{0ex}}\begin{array}{l}\left({H}_{2p}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\frac{1}{2}{H}_{p}\right)\left(2{H}_{2r}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{H}_{r}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{H}_{p+r}\right)\\ \phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\sum _{m=1}^{r-1}\left(\frac{{H}_{m}-{H}_{m+p}}{2m+1}\right)\end{array}\phantom{\rule{0.3em}{0ex}}\right)\phantom{\rule{2em}{0ex}}\end{array}$
(38)

### Proof

To prove Equation 14, we may write

$\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{p}\frac{{h}_{n}\phantom{\rule{1em}{0ex}}}{n\left(\begin{array}{l}n+k\\ k\end{array}\right)}=\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{p}\frac{k!\phantom{\rule{1em}{0ex}}{h}_{n}\phantom{\rule{1em}{0ex}}}{n}\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}\frac{{A}_{r}\phantom{\rule{1em}{0ex}}}{n+r},$
(39)

where A r is given by Equation 10, and by a rearrangement of sums,

$\begin{array}{ll}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{p}\frac{k!\phantom{\rule{1em}{0ex}}{h}_{n}\phantom{\rule{1em}{0ex}}}{n}\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}\frac{{A}_{r}\phantom{\rule{1em}{0ex}}}{n+r}& =\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}r\left(\begin{array}{l}k\\ r\end{array}\right)\sum _{j\phantom{\rule{0.3em}{0ex}}=1}^{p}\phantom{\rule{1em}{0ex}}\sum _{n\phantom{\rule{0.3em}{0ex}}=j}^{p}\left(\frac{1}{n\left(n+r\right)\left(2j-1\right)}\right)\phantom{\rule{2em}{0ex}}\\ =\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}r\left(\begin{array}{l}k\\ r\end{array}\right)\sum _{j\phantom{\rule{0.3em}{0ex}}=1}^{p}\phantom{\rule{1em}{0ex}}\frac{1}{r\left(2j-1\right)}\left(\begin{array}{l}\psi \left(r+j\right)-\psi \left(j\right)\\ -\left(\psi \left(p+1+j\right)-\psi \left(p+1\right)\right)\end{array}\right)\phantom{\rule{2em}{0ex}}\\ =\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}\left(\begin{array}{l}k\\ r\end{array}\right)\sum _{j\phantom{\rule{0.3em}{0ex}}=1}^{p}\phantom{\rule{1em}{0ex}}\frac{1}{\left(2j-1\right)}\left(\begin{array}{l}\frac{1}{j}+\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{1}{m+j}\\ -\frac{1}{p+1}+\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\frac{1}{m+p+1}\end{array}\right)\phantom{\rule{2em}{0ex}}\\ =\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}\left(\begin{array}{l}k\\ r\end{array}\right)\left(\begin{array}{l}2ln2+{H}_{p-\frac{1}{2}}-{H}_{p}-\frac{1}{p+1}\left(ln2+\frac{1}{2}{H}_{p-\frac{1}{2}}\right)\\ +\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\phantom{\rule{1em}{0ex}}\left(\begin{array}{l}\frac{1}{2m+1}\left(2ln2+{H}_{m}+{H}_{p-\frac{1}{2}}-{H}_{m+p}\right)\\ -\frac{1}{p+m+1}\left(ln2+{H}_{p-\frac{1}{2}}\right)\end{array}\right)\end{array}\right)\phantom{\rule{2em}{0ex}}\\ =2ln2+{H}_{p-\frac{1}{2}}-{H}_{p}-\frac{1}{p+1}\left(ln2+\frac{1}{2}{H}_{p-\frac{1}{2}}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}\left(\begin{array}{l}k\\ r\end{array}\right)\left(\begin{array}{l}\left(2ln2+{H}_{p-\frac{1}{2}}\right)\left(-1+ln2+\frac{1}{2}{H}_{r-\frac{1}{2}}\right)\\ -\left(ln2+\frac{1}{2}{H}_{p-\frac{1}{2}}\right)\left({H}_{p+r}-{H}_{p+1}\right)\\ +\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\phantom{\rule{1em}{0ex}}\left(\frac{1}{2m+1}\left({H}_{m}-{H}_{m+p}\right)\right)\end{array}\right)\phantom{\rule{2em}{0ex}}\\ =ln2\left(2ln2+{H}_{p-\frac{1}{2}}\right)-{H}_{p}-\frac{1}{p+1}\left(ln2+\frac{1}{2}{H}_{p-\frac{1}{2}}\right)+{H}_{p+1}\left(ln2+\frac{1}{2}{H}_{p-\frac{1}{2}}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\sum _{r\phantom{\rule{0.3em}{0ex}}=1}^{k}{\left(-1\right)}^{r+1}\left(\begin{array}{l}k\\ r\end{array}\right)\left(\begin{array}{l}\left({H}_{r-\frac{1}{2}}-{H}_{p+r}\right)\left(ln2+\frac{1}{2}{H}_{p-\frac{1}{2}}\right)\\ +\sum _{m\phantom{\rule{0.3em}{0ex}}=1}^{r-1}\phantom{\rule{1em}{0ex}}\left(\frac{1}{2m+1}\left({H}_{m}-{H}_{m+p}\right)\right)\end{array}\right).\phantom{\rule{2em}{0ex}}\end{array}$
(40)

Substituting Equation 7 into Equation 15 and after simplification, Equation 14 follows. □

### Corollary 2

Let $k,p\in \mathrm{ℕ.}$ Then we obtain

$\begin{array}{ll}\phantom{\rule{-14.0pt}{0ex}}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{p}\frac{{H}_{n-\frac{1}{2}}\phantom{\rule{1em}{0ex}}}{n\left(\begin{array}{l}n+k\\ k\end{array}\right)}& =2\sum _{r=1}^{k}{\left(-1\right)}^{r+1}\left(\begin{array}{l}k\\ r\end{array}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\left(\phantom{\rule{0.3em}{0ex}}\begin{array}{l}\left({H}_{2p}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\frac{1}{2}{H}_{p}\right)\left(2{H}_{2r}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{H}_{r}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{H}_{p+r}\right)\\ +\sum _{m=1}^{r-1}\left(\frac{{H}_{m}-{H}_{m+p}}{2m+1}\right)\end{array}\phantom{\rule{0.3em}{0ex}}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+2{H}_{p}\left({H}_{2p}-\frac{1}{2}{H}_{p}-1\right)-\frac{2ln2}{k}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×\left(1-\frac{1\phantom{\rule{1em}{0ex}}}{\left(\begin{array}{l}p+k\\ p\end{array}\right)}\right)\phantom{\rule{2em}{0ex}}\end{array}$
(41)

and

$\begin{array}{ll}\phantom{\rule{-14.0pt}{0ex}}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{p}\frac{{H}_{2n}\phantom{\rule{1em}{0ex}}}{n\left(\begin{array}{l}n+k\\ k\end{array}\right)}\phantom{\rule{0.3em}{0ex}}& ={H}_{p}\left({H}_{2p}-1\right)+\frac{1}{2}{H}_{p}^{\left(2\right)}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{0.3em}{0ex}}+\sum _{r=1}^{k}{\left(-1\right)}^{r+1}\left(\begin{array}{l}k\\ r\end{array}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{0.3em}{0ex}}×\left(\phantom{\rule{0.3em}{0ex}}\begin{array}{l}\left(\phantom{\rule{0.3em}{0ex}}{H}_{2p}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\frac{1}{2}{H}_{p}\right)\phantom{\rule{0.3em}{0ex}}\left(2{H}_{2r}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{H}_{r}-{H}_{p+r}\right)\\ \phantom{\rule{0.3em}{0ex}}+\frac{{H}_{p}}{2}\phantom{\rule{0.3em}{0ex}}\left({H}_{p+r}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{H}_{r-1}\right)\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\sum _{m=1}^{r-1}\phantom{\rule{0.3em}{0ex}}\frac{\left({H}_{m}-{H}_{m+p}\right)}{2m\left(2m+1\right)}\end{array}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\right)\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}\end{array}$
(42)

### Proof

It is straightforward to show that

$\begin{array}{l}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{p}\frac{ln2\phantom{\rule{1em}{0ex}}}{n\left(\begin{array}{l}n+k\\ k\end{array}\right)}=\frac{ln2}{k}\left(1-\frac{1\phantom{\rule{1em}{0ex}}}{\left(\begin{array}{l}p+k\\ p\end{array}\right)}\right);\end{array}$
(43)

then rearranging Equation 14 and using Equation 18, we obtain Equation 16. Rearranging Equation 14 and using Equation 2, we obtain Equation 17. □

### Example 2

Some examples are

$\begin{array}{ll}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{p}\frac{{H}_{2n}-\frac{1}{2}\phantom{\rule{1em}{0ex}}{H}_{n}\phantom{\rule{1em}{0ex}}}{n\left(n+1\right)}& =\frac{2p+1}{p+1}{H}_{2p}-\frac{4p+3}{2\left(p+1\right)}{H}_{p},\phantom{\rule{2em}{0ex}}\\ \sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{p}\frac{{H}_{2n}\phantom{\rule{1em}{0ex}}}{n\left(n+1\right)}& =\frac{2p+1}{p+1}{H}_{2p}-2{H}_{p}+\frac{{H}_{p}^{\left(2\right)}}{2},\phantom{\rule{2em}{0ex}}\end{array}$
(44)
$\begin{array}{ll}\phantom{\rule{-30.0pt}{0ex}}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{p}\frac{{H}_{2n}-\frac{1}{2}\phantom{\rule{1em}{0ex}}{H}_{n}}{n\left(\begin{array}{l}n+2\\ 2\end{array}\right)}& =\frac{\left(2p+1\right)\left(2p+5\right)}{3\left(p+1\right)\left(p+2\right)}{H}_{2p}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\frac{\left(8{p}^{2}+24p+13\right)}{6\left(p+1\right)\left(p+2\right)}{H}_{p}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\frac{p}{3\left(p+1\right)}\phantom{\rule{0.3em}{0ex}},\phantom{\rule{2em}{0ex}}\\ \sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{p}\frac{{H}_{n-\frac{1}{2}}\phantom{\rule{1em}{0ex}}}{n\left(\begin{array}{l}n+2\\ 2\end{array}\right)}& =\frac{\left(2p+1\right)\left(2p+5\right)}{3\left(p+1\right)\left(p+2\right)}{H}_{2p}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\frac{\left(8{p}^{2}+24p+13\right)}{6\left(p+1\right)\left(p+2\right)}{H}_{p}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\frac{p}{3\left(p+1\right)}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\frac{p\left(p+3\right)}{\left(p+1\right)\left(p+2\right)}ln2,\phantom{\rule{2em}{0ex}}\\ \sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{p}\frac{{H}_{2n}}{n\left(\begin{array}{l}n+2\\ 2\end{array}\right)}& =\frac{\left(2p+1\right)\left(2p+5\right)}{3\left(p+1\right)\left(p+2\right)}{H}_{2p}+\frac{{H}_{p}^{\left(2\right)}}{2}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\frac{4{H}_{p}}{3}-\frac{5p}{6\left(p+1\right)}\phantom{\rule{2em}{0ex}}\end{array}$
(45)

and

$\begin{array}{ll}\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{p}\frac{{H}_{2n}-\frac{1}{2}\phantom{\rule{1em}{0ex}}{H}_{n}\phantom{\rule{1em}{0ex}}}{n\left(\begin{array}{l}n+3\\ 3\end{array}\right)}& =\frac{2\left(2p+1\right)\left(4{p}^{2}+22p+33\right)}{15\left(p+1\right)\left(p+2\right)\left(p+3\right)}{H}_{2p}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\frac{\left(16{p}^{3}+96{p}^{2}+176p+81\right)}{15\left(p+1\right)\left(p+2\right)\left(p+3\right)}{H}_{p}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\frac{p\left(11p+25\right)}{30\left(p+1\right)\left(p+2\right)}.\phantom{\rule{2em}{0ex}}\end{array}$
(46)

## Conclusions

The author has generalized some results on variant Euler sums and specifically obtained identities for $\sum _{n\phantom{\rule{0.3em}{0ex}}=1}^{\infty }\frac{{H}_{2n}\phantom{\rule{1em}{0ex}}}{n\left(\begin{array}{l}n+k\\ k\end{array}\right)}$and its finite counterpart.

## Methods

Analytical techniques have been employed in the analysis of our results. We have used many relations of the polygamma functions together with results of reordering of double sums and partial fraction decomposition.

## Author’s information

Professor Anthony Sofo is a Fellow of the Australian Mathematical Society.