New Harmonic Number Series

1. Introduction

Our main purpose in this note is to discover the harmonic number series associated with the following identity:

$$\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^2\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}=\zeta \left(2\right)-H_z^{\left(2\right)},z\in \mathbb{C}\setminus Z^{-},$$

(1)

where $\zeta \left(s\right)$ is the Riemann zeta function and $H_z^{\left(2\right)}$ is a second order harmonic number (both definitions are given below). At $z=0$ identity (1) subsumes the solution to the Basel problem and we will see that its derivative includes the well-known relation between the Apéry constant and a classical Euler sum, namely,

$$\sum _{n=1}^{\infty }{\displaystyle \frac{H_n}{n^2}}=2\zeta \left(3\right),$$

as a special case, $H_j$ being the jth harmonic number.

We will also evaluate the following series:

$$\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n(n+1)\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}},\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n\left(n+1\right)\left(n+2\right)\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}},\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}},$$

and derive the harmonic and odd harmonic number series associated with them.

Identity (1), stated without proof in Sofo and Srivastava [15], is a consequence of the following representation of the digamma function $\psi \left(z\right)={\mathrm{\Gamma }}^{\prime }\left(z\right)/\mathrm{\Gamma }\left(z\right)$:

$$\psi \left(z\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{1}{n+1}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^{k}ln\left(z+k\right),$$

(2)

which holds for all $z\in \mathbb{C}$ with $\Re \left(z\right)>0$. This representation first appeared in [10] and was rediscovered by Boyadzhiev [6].

Harmonic numbers $H_{\alpha }$ and odd harmonic numbers $O_{\alpha }$ are defined for $0\ne \alpha \in \mathbb{C}\setminus {\mathbb{Z}}^{-}$ by the recurrence relations

$$H_{\alpha }=H_{\alpha -1}+{\displaystyle \frac{1}{\alpha }}\mathrm{and}{O}_{\alpha }=O_{\alpha -1}+{\displaystyle \frac{1}{2\alpha -1}},$$

with $H_0=0$ and $O_0=0$. Harmonic numbers are connected to the digamma function through the fundamental relation

$$H_{\alpha }=\psi (\alpha +1)+\gamma .$$

(3)

Generalized harmonic numbers $H_{\alpha }^{\left(m\right)}$ and odd harmonic numbers $O_{\alpha }^{\left(m\right)}$ of order $m\in \mathbb{C}$ are defined by

$$H_{\alpha }^{\left(m\right)}=H_{\alpha -1}^{\left(m\right)}+{\displaystyle \frac{1}{{\alpha }^m}}\mathrm{and}{O}_{\alpha }^{\left(m\right)}=O_{\alpha -1}^{\left(m\right)}+{\displaystyle \frac{1}{{(2\alpha -1)}^m}},$$

with $H_0^{\left(m\right)}=0$ and $O_0^{\left(m\right)}=0$ so that $H_{\alpha }=H_{\alpha }^{\left(1\right)}$ and $O_{\alpha }=O_{\alpha }^{\left(1\right)}$.

The recurrence relations imply that if $\alpha =n$ is a non-negative integer, then

$$H_n^{\left(m\right)}=\sum _{j=1}^n{\displaystyle \frac{1}{j^m}}\mathrm{and}{O}_n^{\left(m\right)}=\sum _{j=1}^n{\displaystyle \frac{1}{{(2j-1)}^m}}.$$

Generalized harmonic numbers are linked to the polygamma functions ${\psi }^{\left(r\right)}\left(z\right)$ of order r defined by

$${\psi }^{\left(r\right)}\left(z\right)={\displaystyle \frac{d^r}{d{z}^r}}\psi \left(z\right)={(-1)}^{r+1}r!\sum _{j=0}^{\infty }{\displaystyle \frac{1}{{(j+z)}^{r+1}}},$$

through

$$H_z^{\left(r\right)}=\zeta \left(r\right)+{\displaystyle \frac{{(-1)}^{r-1}}{(r-1)!}}{\psi }^{(r-1)}(z+1),$$

where $\zeta \left(s\right)$ is the Riemann zeta function defined by

$$\zeta \left(s\right)=\sum _{k=1}^{\infty }{\displaystyle \frac{1}{k^s}},s\in \mathbb{C},\Re \left(s\right)>1.$$

The analytical continuation to all $s\in \mathbb{C}$ with $\Re \left(s\right)>0,s\ne 1,$ is given by

$$\zeta \left(s\right)={(1-2^{1-s})}^{-1}\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k+1}}{k^s}}.$$

2. Proof of Identity (1)

For completeness and a better readability we first prove identity (1).

Theorem 1. 

For all $z\in \mathbb{C}\setminus Z^{-}$ the following identity holds:

$$\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^2\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}=\zeta \left(2\right)-H_z^{\left(2\right)}.$$

Proof. 

Differentiating the representation (2) gives

$$\begin{array}{cc}\hfill {\psi }^{\left(1\right)}\left(z\right)=\zeta \left(2\right)-H_{z-1}^{\left(2\right)}& =\sum _{n=0}^{\infty }{\displaystyle \frac{1}{n+1}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{(-1)}^k}{z+k}}\hfill \\ \hfill & =\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n}}\sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k-1}}\right){\displaystyle \frac{{(-1)}^{k-1}}{z+k-1}}\hfill \\ \hfill & =\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n}}\sum _{k=1}^n{\displaystyle \frac{k}{n}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{(-1)}^{k-1}}{z+k-1}}\hfill \\ \hfill & =\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^2}}\sum _{k=1}^{n}k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{(-1)}^{k-1}}{z+k-1}}\hfill \\ \hfill & =\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^2}}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{n+z-1}{n}\right)}};\hfill \end{array}$$

and hence (1). Note that we used

$$\sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{(-1)}^{k-1}k}{z+k}}={\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}},$$

which is a particular case of the Frisch identity ([1,2]). □

3. Required Identities

We will make frequent use of the following basic identity [12]:

$$\sum _{k=1}^m{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{k+n}{k}\right)}}={\displaystyle \frac{1}{n-1}}-{\displaystyle \frac{n}{n-1}}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{m+n}{m+1}\right)}},0,1\ne n\in \mathbb{C}.$$

(4)

and the identities stated in the following lemmata.

Lemma 1. 

We have

$$\begin{array}{c}{H}_{k-1/2}=2O_k-2ln2,\end{array}$$

(5)

$$\begin{array}{c}{H}_{k-1/2}^{\left(2\right)}=-2\zeta \left(2\right)+4O_k^{\left(2\right)},\end{array}$$

(6)

$$\begin{array}{c}{H}_{-1/2}^{\left(3\right)}=-6\zeta \left(3\right),\end{array}$$

(7)

$$\begin{array}{c}{H}_{-1/2}^{\left(4\right)}=-14\zeta \left(4\right),\end{array}$$

(8)

$$\begin{array}{c}{H}_{k-1/2}^{(m+1)}-H_{-1/2}^{(m+1)}=2^{m+1}{O}_k^{(m+1)}.\end{array}$$

(9)

Lemma 2. 

For integers u and v, we have

$$\begin{array}{c}\left({\displaystyle \genfrac{}{}{0pt}{}{u-1/2}{v}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{2u}{u}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{u}{v}}\right)2^{-2v}{\left({\displaystyle \genfrac{}{}{0pt}{}{2(u-v)}{u-v}}\right)}^{-1},\end{array}$$

(10)

$$\begin{array}{c}\left({\displaystyle \genfrac{}{}{0pt}{}{u}{1/2}}\right)={\displaystyle \frac{2^{2u+1}}{\pi }}{\left({\displaystyle \genfrac{}{}{0pt}{}{2u}{u}}\right)}^{-1},\end{array}$$

(11)

$$\begin{array}{c}\left({\displaystyle \genfrac{}{}{0pt}{}{u}{1/2-v}}\right)={\displaystyle \frac{{(-1)}^{v-1}}{v}}{\left({\displaystyle \genfrac{}{}{0pt}{}{u+v}{v}}\right)}^{-1}{\left({\displaystyle \genfrac{}{}{0pt}{}{2(u+v)}{u+v}}\right)}^{-1}\left({\displaystyle \genfrac{}{}{0pt}{}{2(v-1)}{v-1}}\right)2^{2u+2},\end{array}$$

(12)

$$\begin{array}{c}\left({\displaystyle \genfrac{}{}{0pt}{}{u+1/2}{v}}\right)={\left({\displaystyle \genfrac{}{}{0pt}{}{u}{v}}\right)}^{-1}\left({\displaystyle \genfrac{}{}{0pt}{}{2u+1}{2v}}\right)2^{-2v}\left({\displaystyle \genfrac{}{}{0pt}{}{2v}{v}}\right),\end{array}$$

(13)

$$\begin{array}{c}\left({\displaystyle \genfrac{}{}{0pt}{}{u+1/2}{v}}\right)={\displaystyle \frac{{(-1)}^{v-u-1}}{2^{2v-1}}}{\displaystyle \frac{2u+1}{v-u}}{\left({\displaystyle \genfrac{}{}{0pt}{}{v}{u}}\right)}^{-1}\left({\displaystyle \genfrac{}{}{0pt}{}{2u}{u}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2(v-u-1)}{v-u-1}}\right),v>u,\end{array}$$

(14)

$$\begin{array}{c}\left({\displaystyle \genfrac{}{}{0pt}{}{-3/2}{u}}\right)={(-1)}^u(2u+1)2^{-2u}\left({\displaystyle \genfrac{}{}{0pt}{}{2u}{u}}\right).\end{array}$$

(15)

Proof. 

These identities are readily derived using the following well-known Gamma function identities:

$$\mathrm{\Gamma }\left(u+{\displaystyle \frac{1}{2}}\right)={\displaystyle \frac{\sqrt{\pi }}{2^{2u}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2u}{u}}\right)\mathrm{\Gamma }\left(u+1\right),\mathrm{\Gamma }\left(-u+{\displaystyle \frac{1}{2}}\right)={(-1)}^u{2}^{2u}{\left({\displaystyle \genfrac{}{}{0pt}{}{2u}{u}}\right)}^{-1}{\displaystyle \frac{\sqrt{\pi }}{\mathrm{\Gamma }\left(u+1\right)}},$$

together with the definition of the generalized binomial coefficients:

$$\left({\displaystyle \genfrac{}{}{0pt}{}{u}{v}}\right)={\displaystyle \frac{\mathrm{\Gamma }\left(u+1\right)}{\mathrm{\Gamma }\left(v+1\right)\mathrm{\Gamma }\left(u-v+1\right)}}.$$

□

4. Results

In this section we state new closed forms for infinite series involving inverse factorials. More results of this nature have been produced, among others, by Sofo [13,14], Boyadzhiev [7,8] and by the authors [3].

Theorem 2. 

If m is a non-negative integer, then

$$\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}}{n^2\left(\genfrac{}{}{0pt}{}{2\left(n+m\right)}{n+m}\right)\left(\genfrac{}{}{0pt}{}{n+m}{m}\right)}}={\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}}\left(3\zeta \left(2\right)-4O_m^{\left(2\right)}\right).$$

(16)

In particular,

$$\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}}{n^2\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}={\displaystyle \frac{{\pi }^2}{2}}.$$

(17)

Proof. 

Write $m-1/2$ for z in (1) to obtain

$$\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^2\left(\genfrac{}{}{0pt}{}{n+m-1/2}{n}\right)}}=\zeta \left(2\right)-H_{m-1/2}^{\left(2\right)},$$

(18)

from which (16) follows on account of Lemmata 1 and 2. □

Theorem 3. 

If $z\in \mathbb{C}\setminus {\mathbb{Z}}^{-}$, then

$$\sum _{n=1}^{\infty }{\displaystyle \frac{H_{n+z}}{n^2\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}=H_z\left(\zeta \left(2\right)-H_z^{\left(2\right)}\right)+2\left(\zeta \left(3\right)-H_z^{\left(3\right)}\right).$$

(19)

In particular,

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{H_n}{n^2}}=2\zeta \left(3\right),\end{array}$$

(20)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{H_n}{n^2\left(n+1\right)}}=2\zeta \left(3\right)-\zeta \left(2\right),\end{array}$$

(21)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{H_n}{n^2\left(n+1\right)\left(n+2\right)}}=\zeta \left(3\right)-{\displaystyle \frac{3}{4}}\zeta \left(2\right)+{\displaystyle \frac{1}{4}}.\end{array}$$

(22)

Proof. 

Differentiate (1) with respect to z to obtain

$$\sum _{n=1}^{\infty }{\displaystyle \frac{H_{n+z}-H_z}{n^2\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}=2\left(\zeta \left(3\right)-H_z^{\left(3\right)}\right),$$

(23)

and use (1) again to rewrite the second term on the left hand side of (23).

Identity (20) corresponds to an evaluation of (19) at $z=1$ followed by the use of

$$H_{n+1}=H_n+{\displaystyle \frac{1}{n+1}},$$

(24)

and the decomposition

$${\displaystyle \frac{1}{n^2{\left(n+1\right)}^2}}={\displaystyle \frac{1}{n^2}}+{\displaystyle \frac{1}{{\left(n+1\right)}^2}}-2\left({\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{n+1}}\right).$$

□

Corollary 1. 

If m is a nonnegative integer, then

$$\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}{O}_{n+m}}{n^2\left(\genfrac{}{}{0pt}{}{2\left(n+m\right)}{n+m}\right)\left(\genfrac{}{}{0pt}{}{n+m}{m}\right)}}={\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}}\left(O_m\left(3\zeta \left(2\right)-4O_m^{\left(2\right)}\right)+\left(7\zeta \left(3\right)-8O_m^{\left(3\right)}\right)\right).$$

(25)

Proof. 

Write $m-1/2$ for z in (23) and use Lemmata 1 and 2. □

Theorem 4. 

If $z\in \mathbb{C}\setminus Z^{-}$, then

$$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(H_{n+z}-H_z\right)}^2+H_{n+z}^{\left(2\right)}-H_z^{\left(2\right)}}{n^2\left(\genfrac{}{}{0pt}{}{n+z}{z}\right)}}=6\zeta \left(4\right)-6H_z^{\left(4\right)}.$$

(26)

In particular,

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{H_n^2+H_n^{\left(2\right)}}{n^2}}=6\zeta \left(4\right),\end{array}$$

(27)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^2}}{\displaystyle \frac{2^{2n}}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left(O_n^2+O_n^{\left(2\right)}\right)={\displaystyle \frac{{\pi }^2}{4}}.\end{array}$$

(28)

Proof. 

Differentiate (23) with respect to z. □

Remark 1. 

The symmetry relation

$$\sum _{n=1}^{\infty }{\displaystyle \frac{H_n^{\left(p\right)}}{n^q}}+\sum _{n=1}^{\infty }{\displaystyle \frac{H_n^{\left(q\right)}}{n^p}}=\zeta \left(p\right)\zeta \left(q\right)+\zeta (p+q)$$

makes it easy to calculate

$$\sum _{n=1}^{\infty }{\displaystyle \frac{H_n^{\left(p\right)}}{n^p}}={\displaystyle \frac{1}{2}}\left(\zeta {\left(p\right)}^2+\zeta \left(2p\right)\right),$$

so that, in particular,

$$\sum _{n=1}^{\infty }{\displaystyle \frac{H_n^{\left(2\right)}}{n^2}}={\displaystyle \frac{7}{4}}\zeta \left(4\right);$$

and using this in (27) therefore gives

$$\sum _{n=1}^{\infty }{\displaystyle \frac{H_n^2}{n^2}}={\displaystyle \frac{17}{4}}\zeta \left(4\right).$$

(29)

Thus (27) allows a much easier determination of the sum (29) which was originally evaluated by Borwein and Borwein [5] through Fourier series and contour integration.

Lemma 3. 

If $z\in \mathbb{C}\setminus {\mathbb{Z}}^{-}$, then

$$\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n(n+1)\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}=z\left(H_z^{\left(2\right)}-\zeta \left(2\right)\right)+1,$$

(30)

Proof. 

Sum (1) over z from 1 to r, using (4) and the fact that [4]

$$\sum _{z=1}^r{H}_z^{\left(2\right)}=\left(r+1\right)H_r^{\left(2\right)}-H_r,$$

to obtain

$$\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n(n+1)\left(\genfrac{}{}{0pt}{}{n+r}{r}\right)}}=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{{(n+1)}^{2}n}}-1-(r-1)\zeta \left(2\right)+r{H}_r^{\left(2\right)}.$$

But

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{1}{{(n+1)}^{2}n}}& =\sum _{n=1}^{\infty }\left({\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{n+1}}\right)-\sum _{n=1}^{\infty }{\displaystyle \frac{1}{{(n+1)}^2}}\hfill \\ \hfill & =\sum _{n=1}^{\infty }\left({\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{n+1}}\right)-\sum _{n=2}^{\infty }{\displaystyle \frac{1}{n^2}}\hfill \\ \hfill & =\sum _{n=1}^{\infty }\left({\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{n+1}}\right)+1-\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^2}}\hfill \\ \hfill & =2-\zeta \left(2\right),\hfill \end{array}$$

which when substituted into the previous sum gives (30) after replacing r with z. □

Remark 2. 

Identity (30) was also derived by Sofo and Srivastava [15].

Theorem 5. 

If m is a non-negative integer, then

$$\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}}{n\left(n+1\right)\left(\genfrac{}{}{0pt}{}{2\left(n+m\right)}{n+m}\right)\left(\genfrac{}{}{0pt}{}{n+m}{m}\right)}}={\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}}\left(\left(m-{\displaystyle \frac{1}{2}}\right)\left(4O_m^{\left(2\right)}-3\zeta \left(2\right)\right)+1\right).$$

(31)

In particular,

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}}{n\left(n+1\right)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}={\displaystyle \frac{{\pi }^2}{4}}+1,\end{array}$$

(32)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}}{n{\left(n+1\right)}^2\left(\genfrac{}{}{0pt}{}{2\left(n+1\right)}{n+1}\right)}}={\displaystyle \frac{3}{2}}-{\displaystyle \frac{{\pi }^2}{8}},\end{array}$$

(33)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}}{n{\left(n+1\right)}^2\left(n+2\right)\left(\genfrac{}{}{0pt}{}{2\left(n+2\right)}{n+2}\right)}}={\displaystyle \frac{23}{36}}-{\displaystyle \frac{{\pi }^2}{16}}.\end{array}$$

(34)

Theorem 6. 

If $z\in \mathbb{C}\setminus {\mathbb{Z}}^{-}$, then

$$\sum _{n=1}^{\infty }{\displaystyle \frac{H_{n+z}}{n\left(n+1\right)\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}=\left(\zeta \left(2\right)-H_z^{\left(2\right)}\right)\left(1-z{H}_z\right)+H_z-2z\left(\zeta \left(3\right)-H_z^{\left(3\right)}\right).$$

(35)

In particular,

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{H_n}{n\left(n+1\right)}}={\displaystyle \frac{{\pi }^2}{6}},\end{array}$$

(36)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{H_n}{n{\left(n+1\right)}^2}}=\zeta \left(2\right)-\zeta \left(3\right).\end{array}$$

(37)

Proof. 

Differentiate (30) with respect to z to obtain

$$\sum _{n=1}^{\infty }{\displaystyle \frac{H_{n+z}-H_z}{n(n+1)\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}=\zeta \left(2\right)-H_z^{\left(2\right)}-2z\left(\zeta \left(3\right)-H_z^{\left(3\right)}\right),$$

(38)

and hence (35) upon using (30) again to rewrite the second sum on the left hand side of (38). Identity () is an evaluation of (35) at $z=1$ where we used (24) and the fact that

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{1}{n{(n+1)}^3}}& =\sum _{n=1}^{\infty }\left({\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{n+1}}-{\displaystyle \frac{1}{{(n+1)}^2}}-{\displaystyle \frac{1}{{(n+1)}^3}}\right)\hfill \\ \hfill & =\sum _{n=1}^{\infty }\left({\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{n+1}}\right)-\sum _{n=1}^{\infty }{\displaystyle \frac{1}{{(n+1)}^2}}-\sum _{n=1}^{\infty }{\displaystyle \frac{1}{{(n+1)}^3}}.\hfill \end{array}$$

□

Remark 3. 

Identity (36) was also recorded by Chu in [9].

Remark 4. 

Subtraction of () from (20) gives the Euler sum

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

(39)

Corollary 2. 

If m is a non-negative integer, then

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}{O}_{n+m}}{n\left(n+1\right)\left(\genfrac{}{}{0pt}{}{2\left(n+m\right)}{n+m}\right)\left(\genfrac{}{}{0pt}{}{n+m}{m}\right)}}& ={\displaystyle \frac{1}{2\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}}\left(3\zeta \left(2\right)-4O_m^{\left(2\right)}\right)\left(1-O_m\left(2m-1\right)\right)\hfill \\ \hfill & +{\displaystyle \frac{O_m}{\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}}-{\displaystyle \frac{\left(2m-1\right)}{2\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}}\left(7\zeta \left(3\right)-8O_m^{\left(3\right)}\right).\hfill \end{array}$$

(40)

In particular,

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}{O}_n}{n\left(n+1\right)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}={\displaystyle \frac{{\pi }^2}{4}}+{\displaystyle \frac{7}{2}}\zeta \left(3\right),\end{array}$$

(41)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}{O}_{n+1}}{n{\left(n+1\right)}^2\left(\genfrac{}{}{0pt}{}{2\left(n+1\right)}{n+1}\right)}}={\displaystyle \frac{5}{2}}-{\displaystyle \frac{7}{4}}\zeta \left(3\right).\end{array}$$

(42)

Proof. 

Write $m-1/2$ for z in (38) and use Lemmata 1 and 2. □

Theorem 7. 

If $z\in \mathbb{C}\setminus {\mathbb{Z}}^{-}$, then

$$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(H_{n+z}-H_z\right)}^2+H_{n+z}^{\left(2\right)}-H_z^{\left(2\right)}}{n(n+1)\left(\genfrac{}{}{0pt}{}{n+z}{z}\right)}}=4\left(\zeta \left(3\right)-H_z^{\left(3\right)}\right)-6z\left(\zeta \left(4\right)-H_z^{\left(4\right)}\right).$$

(43)

In particular,

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{H_n^2+H_n^{\left(2\right)}}{n(n+1)}}=4\zeta \left(3\right),\end{array}$$

(44)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}\left(O_n^2+O_n^{\left(2\right)}\right)}{n(n+1)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}={\displaystyle \frac{{\pi }^4}{8}}+7\zeta \left(3\right).\end{array}$$

(45)

Proof. 

Differentiate (38) with respect to z. □

Remark 5. 

Since Xu showed that [17, Equation (2.30)]

$$\sum _{n=1}^{\infty }{\displaystyle \frac{H_n^{\left(2\right)}}{n\left(n+1\right)}}=\zeta \left(3\right),$$

identity (44) yields

$$\sum _{n=1}^{\infty }{\displaystyle \frac{H_n^2}{n\left(n+1\right)}}=3\zeta \left(3\right);$$

(46)

an identity that was also reported by Nimbran and Sofo in [11].

Lemma 4. 

If $z\in \mathbb{C}\setminus {\mathbb{Z}}^{-}$, then,

$$\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n\left(n+1\right)\left(n+2\right)\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}={\displaystyle \frac{1}{2}}z\left(z+1\right)\left(H_z^{\left(2\right)}-\zeta \left(2\right)\right)+{\displaystyle \frac{z}{2}}+{\displaystyle \frac{1}{4}}.$$

(47)

Proof. 

We first derive the following identity:

$$\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n\left(n+2\right)\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}={\displaystyle \frac{1}{2}}z\left(z-1\right)\left(\zeta \left(2\right)-H_z^{\left(2\right)}\right)-{\displaystyle \frac{z}{2}}+{\displaystyle \frac{3}{4}},$$

(48)

by summing both sides of (30) over z from 1 to r and replacing r with z in the final identity. Note that

$$\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n\left(n+1\right)\left(n+2\right)}}={\displaystyle \frac{1}{4}},$$

since

$${\displaystyle \frac{1}{n\left(n+1\right)\left(n+2\right)}}={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{n+1}}-{\displaystyle \frac{1}{n+1}}+{\displaystyle \frac{1}{n+2}}\right).$$

Note also that [4, Equation (3.2)]:

$$2\sum _{z=1}^{r}z{H}_z^{\left(2\right)}=r\left(r+1\right)H_r^{\left(2\right)}+H_r-r.$$

(49)

Identity (47) is obtained by subtracting (48) from (30). □

Theorem 8. 

If m is a non-negative integer, then

$$\begin{array}{cc}\hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}}{n\left(n+1\right)\left(n+2\right)\left(\genfrac{}{}{0pt}{}{2\left(n+m\right)}{n+m}\right)\left(\genfrac{}{}{0pt}{}{n+m}{m}\right)}}\hfill \\ \hfill & ={\displaystyle \frac{1}{2\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}}\left(\left(m-{\displaystyle \frac{1}{2}}\right)\left(m+{\displaystyle \frac{1}{2}}\right)\left(4O_m^{\left(2\right)}-3\zeta \left(2\right)\right)+m\right).\hfill \end{array}$$

(50)

In particular,

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}}{n\left(n+1\right)\left(n+2\right)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}={\displaystyle \frac{{\pi }^2}{16}},\end{array}$$

(51)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}}{n{\left(n+1\right)}^2\left(n+2\right)\left(\genfrac{}{}{0pt}{}{2\left(n+1\right)}{n+1}\right)}}=1-{\displaystyle \frac{3{\pi }^2}{32}},\end{array}$$

(52)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n+1}}{n{\left(n+1\right)}^2{\left(n+2\right)}^2\left(\genfrac{}{}{0pt}{}{2\left(n+2\right)}{n+2}\right)}}={\displaystyle \frac{14}{9}}-{\displaystyle \frac{5{\pi }^2}{32}}.\end{array}$$

(53)

Proof. 

Set $z=m-1/2$ in (47). □

Theorem 9. 

If $z\in \mathbb{C}\setminus {\mathbb{Z}}^{-}$, then

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{H_{n+z}}{n\left(n+1\right)\left(n+2\right)\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}& =\left(z+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2}}z\left(z+1\right)H_z\right)\left(\zeta \left(2\right)-H_z^{\left(2\right)}\right)\hfill \\ \hfill & +\left({\displaystyle \frac{z}{2}}+{\displaystyle \frac{1}{4}}\right)H_z-z\left(z+1\right)\left(\zeta \left(3\right)-H_z^{\left(3\right)}\right)-{\displaystyle \frac{1}{2}}.\hfill \end{array}$$

(54)

In particular,

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{H_n}{n\left(n+1\right)\left(n+2\right)}}={\displaystyle \frac{{\pi }^2}{12}}-{\displaystyle \frac{1}{2}},\end{array}$$

(55)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{H_n}{n{\left(n+1\right)}^2\left(n+2\right)}}={\displaystyle \frac{{\pi }^2}{12}}+{\displaystyle \frac{1}{2}}-\zeta \left(3\right).\end{array}$$

(56)

Proof. 

Differentiate (47) to obtain

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{H_{n+z}-H_z}{n\left(n+1\right)\left(n+2\right)\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}& =\left(z+{\displaystyle \frac{1}{2}}\right)\left(\zeta \left(2\right)-H_z^{\left(2\right)}\right)\hfill \\ \hfill & -z\left(z+1\right)\left(\zeta \left(3\right)-H_z^{\left(3\right)}\right)-{\displaystyle \frac{1}{2}}.\hfill \end{array}$$

(57)

Identity (55) is an evaluation of (54) at $z=1$, where we also used (24) and the fact that

$$\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n{\left(n+1\right)}^3\left(n+2\right)}}={\displaystyle \frac{5}{4}}-\zeta \left(3\right),$$

since

$${\displaystyle \frac{1}{n{\left(n+1\right)}^3\left(n+2\right)}}={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{n+1}}\right)-{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1}{n+1}}-{\displaystyle \frac{1}{n+2}}\right)-{\displaystyle \frac{1}{{\left(n+1\right)}^3}}.$$

□

Theorem 10. 

If m is a non-negative integer, then

$$\begin{array}{cc}\hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}{O}_{n+m}}{n\left(n+1\right)\left(n+2\right)\left(\genfrac{}{}{0pt}{}{2\left(n+m\right)}{n+m}\right)\left(\genfrac{}{}{0pt}{}{n+m}{m}\right)}}\hfill \\ \hfill & ={\displaystyle \frac{1}{2\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}}\left(m-O_m\left(m-{\displaystyle \frac{1}{2}}\right)\left(m+{\displaystyle \frac{1}{2}}\right)\right)\left(3\zeta \left(2\right)-4O_m^{\left(2\right)}\right)\hfill \\ \hfill & -{\displaystyle \frac{1}{2\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}}\left(\left(m-{\displaystyle \frac{1}{2}}\right)\left(m+{\displaystyle \frac{1}{2}}\right)\left(7\zeta \left(3\right)-8O_m^{\left(3\right)}\right)-m{O}_m+{\displaystyle \frac{1}{2}}\right).\hfill \end{array}$$

(58)

In particular,

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}{O}_n}{n\left(n+1\right)\left(n+2\right)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}={\displaystyle \frac{7}{8}}\zeta \left(3\right)-{\displaystyle \frac{1}{4}},\end{array}$$

(59)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}{O}_{n+1}}{n{\left(n+1\right)}^2\left(n+2\right)\left(\genfrac{}{}{0pt}{}{2\left(n+1\right)}{n+1}\right)}}={\displaystyle \frac{{\pi }^2}{32}}-{\displaystyle \frac{21}{16}}\zeta \left(3\right)+{\displaystyle \frac{11}{8}}.\end{array}$$

(60)

Proof. 

Set $z=m-1/2$ in (57) and use Lemmata 1 and 2. □

Theorem 11. 

If $z\in \mathbb{C}\setminus {\mathbb{Z}}^{-}$, then

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{{\left(H_{n+z}-H_z\right)}^2+H_{n+z}^{\left(2\right)}-H_z^{\left(2\right)}}{n\left(n+1\right)\left(n+2\right)\left(\genfrac{}{}{0pt}{}{n+z}{z}\right)}}& =H_z^{\left(2\right)}-\zeta \left(2\right)-2\left(2z+1\right)\left(H_z^{\left(3\right)}-\zeta \left(3\right)\right)\hfill \\ \hfill & +3z\left(z+1\right)\left(H_z^{\left(4\right)}-\zeta \left(4\right)\right).\hfill \end{array}$$

(61)

In particular,

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{H_n^2+H_n^{\left(2\right)}}{n\left(n+1\right)\left(n+2\right)}}=2\zeta \left(3\right)-\zeta \left(2\right),\end{array}$$

(62)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}\left(O_n^2+O_n^{\left(2\right)}\right)}{n\left(n+1\right)\left(n+2\right)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}={\displaystyle \frac{{\pi }^4}{32}}-{\displaystyle \frac{{\pi }^2}{8}}.\end{array}$$

(63)

Proof. 

Differentiate (57) with respect to z. □

Lemma 5. 

If $z\in \mathbb{C}\setminus {\mathbb{Z}}^{-}$, then

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}& ={\displaystyle \frac{1}{12}}z\left(z+1\right)\left(z+2\right)\left(H_z^{\left(2\right)}-\zeta \left(2\right)\right)\hfill \\ \hfill & +{\displaystyle \frac{z^2}{12}}+{\displaystyle \frac{5z}{24}}+{\displaystyle \frac{1}{18}}.\hfill \end{array}$$

(64)

Proof. 

Sum (48) over z using (4), (49) and [4]

$$\sum _{z=1}^r{z}^2{H}_z^{\left(2\right)}={\displaystyle \frac{r\left(r+1\right)\left(2r+1\right)}{6}}{H}_r^{\left(2\right)}-{\displaystyle \frac{1}{6}}{H}_r+{\displaystyle \frac{r}{3}}-{\displaystyle \frac{r^2}{6}},$$

to obtain

$$\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n\left(n+3\right)\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}={\displaystyle \frac{1}{6}}z\left(z-1\right)\left(z-2\right)\left(H_z^{\left(2\right)}-\zeta \left(2\right)\right)+{\displaystyle \frac{1}{6}}{z}^2-{\displaystyle \frac{7}{12}}z+{\displaystyle \frac{11}{18}}.$$

(65)

Identity (64) is obtained by adding (65) and (47) and subtracting (48). □

Theorem 12. 

If m is a non-negative integer, then

$$\begin{array}{cc}\hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(\genfrac{}{}{0pt}{}{2\left(n+m\right)}{n+m}\right)\left(\genfrac{}{}{0pt}{}{n+m}{m}\right)}}\hfill \\ \hfill & ={\displaystyle \frac{1}{12\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}}\left(m-{\displaystyle \frac{1}{2}}\right)\left(m+{\displaystyle \frac{1}{2}}\right)\left(m+{\displaystyle \frac{3}{2}}\right)\left(4O_m^{\left(2\right)}-3\zeta \left(2\right)\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{4\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}}\left({\displaystyle \frac{m^2}{3}}+{\displaystyle \frac{m}{2}}-{\displaystyle \frac{1}{9}}\right).\hfill \end{array}$$

(66)

In particular,

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}={\displaystyle \frac{{\pi }^2}{64}}-{\displaystyle \frac{1}{36}},\end{array}$$

(67)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}}{n{\left(n+1\right)}^2\left(n+2\right)\left(n+3\right)\left(\genfrac{}{}{0pt}{}{2\left(n+1\right)}{n+1}\right)}}={\displaystyle \frac{29}{72}}-{\displaystyle \frac{5{\pi }^2}{128}},\end{array}$$

(68)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n+1}}{n{\left(n+1\right)}^2{\left(n+2\right)}^2\left(n+3\right)\left(\genfrac{}{}{0pt}{}{2\left(n+2\right)}{n+2}\right)}}={\displaystyle \frac{65}{72}}-{\displaystyle \frac{35{\pi }^2}{384}}.\end{array}$$

(69)

Proof. 

Set $z=m-1/2$ in (64). □

Theorem 13. 

If $z\in \mathbb{C}\setminus {\mathbb{Z}}^{-}$, then

$$\begin{array}{cc}\hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{H_{n+z}}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}\hfill \\ \hfill & =\left({\displaystyle \frac{1}{12}}z\left(z+1\right)\left(z+2\right)H_z-{\displaystyle \frac{z\left(z+2\right)}{4}}-{\displaystyle \frac{1}{6}}\right)\left(H_z^{\left(2\right)}-\zeta \left(2\right)\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{6}}z\left(z+1\right)\left(z+2\right)\left(H_z^{\left(3\right)}-\zeta \left(3\right)\right)+\left({\displaystyle \frac{z^2}{12}}+{\displaystyle \frac{5z}{24}}+{\displaystyle \frac{1}{18}}\right)H_z-{\displaystyle \frac{z}{6}}-{\displaystyle \frac{5}{24}}.\hfill \end{array}$$

(70)

In particular,

$$\sum _{n=1}^{\infty }{\displaystyle \frac{H_n}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}}={\displaystyle \frac{{\pi }^2}{36}}-{\displaystyle \frac{5}{24}}.$$

(71)

Proof. 

Differentiate (64) to obtain

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{H_{n+z}-H_z}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(\genfrac{}{}{0pt}{}{n+z}{n}\right)}}& =-\left({\displaystyle \frac{z\left(z+2\right)}{4}}+{\displaystyle \frac{1}{6}}\right)\left(H_z^{\left(2\right)}-\zeta \left(2\right)\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{6}}z\left(z+1\right)\left(z+2\right)\left(H_z^{\left(3\right)}-\zeta \left(3\right)\right)-{\displaystyle \frac{z}{6}}-{\displaystyle \frac{5}{24}},\hfill \end{array}$$

(72)

and hence (70) by a repeated use of (64). □

Theorem 14. 

If m is a non-negative integer, then

$$\begin{array}{cc}\hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}{O}_{n+m}}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(\genfrac{}{}{0pt}{}{2\left(n+m\right)}{n+m}\right)\left(\genfrac{}{}{0pt}{}{n+m}{m}\right)}}\hfill \\ \hfill & ={\displaystyle \frac{1}{2\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}}\left({\displaystyle \frac{1}{6}}\left(m-{\displaystyle \frac{1}{2}}\right)\left(m+{\displaystyle \frac{1}{2}}\right)\left(m+{\displaystyle \frac{3}{2}}\right)O_m-{\displaystyle \frac{m^2}{4}}-{\displaystyle \frac{m}{4}}+{\displaystyle \frac{1}{48}}\right)\left(4O_m^{\left(2\right)}-3\zeta \left(2\right)\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{12\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}}\left(m-{\displaystyle \frac{1}{2}}\right)\left(m+{\displaystyle \frac{1}{2}}\right)\left(m+{\displaystyle \frac{3}{2}}\right)\left(8O_m^{\left(3\right)}-7\zeta \left(3\right)\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}}\left({\displaystyle \frac{1}{36}}\left(6m^2+9m-2\right)O_m-{\displaystyle \frac{m}{6}}-{\displaystyle \frac{1}{8}}\right).\hfill \end{array}$$

(73)

In particular,

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}{O}_n}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}={\displaystyle \frac{7}{32}}\zeta \left(3\right)-{\displaystyle \frac{{\pi }^2}{192}}-{\displaystyle \frac{1}{16}},\end{array}$$

(74)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}{O}_{n+1}}{n{\left(n+1\right)}^2\left(n+2\right)\left(n+3\right)\left(\genfrac{}{}{0pt}{}{2\left(n+1\right)}{n+1}\right)}}={\displaystyle \frac{137}{288}}+{\displaystyle \frac{{\pi }^2}{48}}-{\displaystyle \frac{35}{64}}\zeta \left(3\right).\end{array}$$

(75)

Proof. 

Write $m-1/2$ for z in (72) and use Lemmata 1 and 2. □

Theorem 15. 

If $z\in \mathbb{C}\setminus {\mathbb{Z}}^{-}$, then

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{{\left(H_{n+z}-H_z\right)}^2+H_{n+z}^{\left(2\right)}-H_z^{\left(2\right)}}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(\genfrac{}{}{0pt}{}{n+z}{z}\right)}}& ={\displaystyle \frac{1}{2}}\left(z+1\right)\left(H_z^{\left(2\right)}-\zeta \left(2\right)\right)\hfill \\ \hfill & -\left(z\left(z+2\right)+{\displaystyle \frac{2}{3}}\right)\left(H_z^{\left(3\right)}-\zeta \left(3\right)\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}z\left(z+1\right)\left(z+2\right)\left(H_z^{\left(4\right)}-\zeta \left(4\right)\right)+{\displaystyle \frac{1}{6}}.\hfill \end{array}$$

(76)

In particular,

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{H_n^2+H_n^{\left(2\right)}}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}}={\displaystyle \frac{2}{3}}\zeta \left(3\right)-{\displaystyle \frac{{\pi }^2}{12}}+{\displaystyle \frac{1}{6}},\end{array}$$

(77)

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n}\left(O_n^2+O_n^{\left(2\right)}\right)}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}={\displaystyle \frac{{\pi }^4}{128}}-{\displaystyle \frac{{\pi }^2}{32}}-{\displaystyle \frac{7}{48}}\zeta \left(3\right)+{\displaystyle \frac{1}{24}}.\end{array}$$

(78)

Proof. 

Differentiate (72) with respect to z. □