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# Harmonic number series associated with certain generating functions

Kunle Adegoke, Robert Frontczak, Taras Goy

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Kunle Adegoke, Robert Frontczak, Taras Goy. Harmonic number series associated with certain generating functions. 2025. hal-05111656

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# Harmonic number series associated with certain generating functions

Kunle Adegoke & Robert Frontczak & Taras Goy

**Abstract.** We present several series involving central binomial coefficients, Catalan numbers $C_n$, harmonic numbers $H_n$, odd harmonic numbers $O_n$, and various products of these classical sequences. A central highlight of our analysis is the derivation of closed-form expressions for two notable classes of series: $\sum_{n=1}^{\infty} \binom{2n}{n} H_n O_n x^n$ and $\sum_{n=1}^{\infty} C_n H_n O_n x^n$. Our results build upon and complement the work of Lehmer, Boyadzhiev, Chen, Li and Chu, Sofo, and several other researchers in this area of analytic and combinatorial number theory. In addition, we introduce new families of series involving products of harmonic and related numbers, some of which yield elegant and unexpected identities. Some of our series cannot be expressed using elementary functions and instead involve combinations of dilogarithm functions. Only in exceptional cases–such as when the golden ratio appears–can they be written in simple closed forms. These findings shed new light on the rich connections between combinatorics and classical analysis.

**Keywords.** Series, harmonic number, odd harmonic number, central binomial coefficient, Catalan number, dilogarithm, golden ratio.

**2010 Mathematics Subject Classification.** 33B15, 11B39

## 1. Introduction and motivation

The central binomial coefficients are defined for nonnegative $n$ by $\binom{2n}{n} = \frac{(2n)!}{(n!)^2}$. These numbers are closely related to the famous Catalan numbers $C_n = \frac{1}{n+1}\binom{2n}{n}$, which can also be expressed by the recursion

$$C_n = \frac{2(2n-1)}{n+1}C_{n-1}, \quad C_0 = 1. \qquad (1.1)$$

Catalan numbers have a long history, possess beautiful properties, and play an outstanding role in combinatorics. Excellent sources on these numbers are the books by Koshy [Kos09], Roman [Rom15] and Stanley [Sta15]. Some examples of recent work involving Catalan numbers include [AFG22, AFG25, Alz20, Chu21, GoSh19, Mik19, QiGu17, ZhCh19]. The ordinary generating functions of the numbers $\binom{2n}{n}$ and $C_n$ are well known [Leh85]:

$$\sum_{n=0}^{\infty} \binom{2n}{n} x^n = \frac{1}{\sqrt{1-4x}} \quad \text{and} \quad \sum_{n=0}^{\infty} C_n x^n = \frac{2}{1+\sqrt{1-4x}}. \qquad (1.2)$$

Both series converge for $|x| < 1/4$.

We proceed with some definitions and basic statements. Let $\text{Li}_n(x)$ and $\zeta(s)$ denote the polylogarithm functions and the Riemann zeta function, respectively. These functions are defined by [Lew81, SrCh01]

$$\text{Li}_0(x) = \frac{x}{1-x}, \quad \text{Li}_1(x) = -\ln(1-x), \quad \text{and} \quad \text{Li}_n(x) = \sum_{k=1}^{\infty} \frac{x^k}{k^n}, \quad |x| \le 1, n \ge 2,$$

$$\zeta(s) = \begin{cases} \sum_{k=1}^{\infty} \frac{1}{k^s} = \frac{1}{1-2^{-s}} \sum_{k=1}^{\infty} \frac{1}{(2k-1)^s}, & \Re(s) > 1, \\ \frac{1}{1-2^{1-s}} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^s}, & \Re(s) > 0; s \ne 1. \end{cases}$$
For $n \ge 2$, we have the identities $\mathrm{Li}_n(1) = \zeta(n)$ and $\mathrm{Li}_n(-1) = -(1 - 2^{1-n})\zeta(n)$. The function $\mathrm{Li}_2(x)$ is the familiar dilogarithm, and we have

$$ \mathrm{Li}_2\left(\frac{1}{2}\right) = \frac{\pi^2}{12} - \frac{1}{2}\ln^2 2. \qquad (1.3) $$

Among the many beautiful properties of the dilogarithm, the following identity will be used frequently in the paper:

$$ \mathrm{Li}_2(y) = -\mathrm{Li}_2\left(\frac{1}{y}\right) + \frac{\pi^2}{3} - \frac{1}{2}\ln^2 y - i\pi \ln y, \quad y > 1, i = \sqrt{-1}. \qquad (1.4) $$

Harmonic numbers of order $m$, denoted $H_n^{(m)}$, and odd harmonic numbers of order $m$, denoted $O_n^{(m)}$, are defined by

$$ H_n^{(m)} = \sum_{k=1}^{n} \frac{1}{k^m} \quad \text{and} \quad O_n^{(m)} = \sum_{k=1}^{n} \frac{1}{(2k-1)^m}, $$

where $H_0^{(m)} = O_0^{(m)} = 0$, and $H_n^{(1)} = H_n$ and $O_n^{(1)} = O_n$ are the ordinary harmonic and odd harmonic numbers, respectively. Harmonic numbers and odd harmonic numbers are related, among other ways, by

$$ H_{2n} = \frac{1}{2}H_n + O_n \quad \text{and} \quad H_{2n-1} = \frac{1}{2}H_{n-1} + O_n. $$

A fundamental connection between harmonic numbers and the digamma function $\psi(z) = \frac{\Gamma'(z)}{\Gamma(z)}$ is given by [SrCh01]

$$ H_z = \psi(z+1) + \gamma, \qquad (1.5) $$

with $\gamma$ being the Euler-Mascheroni constant and $\psi(z)$ equals

$$ \psi(z) = -\gamma + \sum_{k=0}^{\infty} \left( \frac{1}{k+1} - \frac{1}{k+z} \right). $$

Similarly, generalized harmonic numbers $H_n^{(m)}$ are connected to the polygamma functions $\psi^{(m)}(z)$ of order $m$, defined by

$$ \psi^{(m)}(z) = \frac{d^m}{dz^m} \psi(z) = (-1)^{m+1} m! \sum_{j=0}^{\infty} \frac{1}{(j+z)^{m+1}}, $$

through the relation

$$ H_z^{(m)} = \zeta(m) + \frac{(-1)^{m-1}}{(m-1)!} \psi^{(m-1)}(z+1). \qquad (1.6) $$

These connections imply that generalized harmonic numbers at rational arguments $H_{p/q}^{(m)}$ must be interpreted as polygamma functions evaluated at rational arguments. As shown by Kölbig [Kol96], and later by Choi and Cvijović [ChCv07], the polygamma function $\psi^{(m-1)}(p/q + 1)$ can be explicitly expressed in terms of polylogarithms or other special functions. For instance, we have

$$ H_{-3/4} = -\frac{\pi}{2} - 3 \ln 2 \quad \text{and} \quad H_{-1/6}^{(3)} = -2\sqrt{3\pi^3} - 90 \zeta(3). $$

The correspondence between $H_z^{(m)}$ and $\psi^{(m)}(z)$ also enables the derivation of further relations between harmonic numbers and odd harmonic numbers. Two such relations relevant to this study are

$$ H_{n-1/2} = 2O_n - 2 \ln 2 \quad \text{and} \quad H_{n-1/2} - H_{-1/2} = 2O_n. \qquad (1.7) $$
This paper is particularly motivated by the papers by Boyadzhiev [Boy12] and Chen [Ch16]. Boyadzhiev studied series (generating functions) involving central binomial coefficients and harmonic numbers $H_n$. His main results include [Boy12, Theorem 1]

$$ \sum_{n=0}^{\infty} \binom{2n}{n} H_n x^n = \frac{2}{\sqrt{1-4x}} \ln \left( \frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}} \right), \quad x \in [-1/4; 1/4], \qquad (1.8) $$

and the generating function for the product $C_n H_n$, valid for $|x| \le 1/4$ [Boy12, Corollary 2]

$$ \sum_{n=0}^{\infty} C_n H_n x^n = \frac{1}{x} \left( \sqrt{1-4x} \ln(2\sqrt{1-4x}) - (1+\sqrt{1-4x}) \ln(1+\sqrt{1-4x}) + \ln 2 \right). \qquad (1.9) $$

Chen added results for the sequences $\binom{2n}{n}(H_{2n}-H_n)$, $C_n(H_{2n}-H_n)$, and $\binom{2n}{n}O_n$. Among other results, he proved the identities [Ch16, Theorem 8]

$$ \sum_{n=1}^{\infty} \binom{2n}{n} O_n x^n = - \frac{\ln(1-4x)}{2\sqrt{1-4x}}, \quad x \in [-1/4; 1/4], \qquad (1.10) $$

as well as [Ch16, Corollary 9]

$$ \sum_{n=1}^{\infty} C_n O_n x^n = \frac{1}{2x} \left( 1 - \sqrt{1-4x} + \frac{1}{2} \sqrt{1-4x} \ln(1-4x) \right), \quad |x| < 1/4. \qquad (1.11) $$

See also [AFG25] for alternative proofs of Chen's results, as well as generalizations and related combinatorial identities.

The goal of the present paper is twofold. First, we derive and evaluate in closed form a range of harmonic number series associated with certain known generating functions. Second, we establish new expressions for two classes of series involving products of harmonic numbers $H_n$ and odd harmonic numbers $O_n$, respectively. These classes are $\sum_{n=1}^{\infty} \binom{2n}{n} H_n O_n x^n$ and $\sum_{n=1}^{\infty} C_n H_n O_n x^n$. The method employed to prove these expressions is well known and is based on integration. The evaluation of infinite series via integrals is a classical analytical technique, recently revitalized in works by Xu [Xu17], Sofo and Nimbran [SoNi19], Stewart [Ste20], and Li and Chu [LiCh22-a, LiCh22-b, LiCh22-c, LiCh24]. We also derive new series involving products of harmonic numbers.

To provide the reader with a sense of the results to come, we present the following example:

$$ \sum_{n=1}^{\infty} \frac{H_n H_{n+r+s+1}}{\binom{n+r+s+1}{r+2}} = \frac{r+2}{s \binom{r+s}{r}} \left( (H_{r+s} - H_r) \left(H_{r+s} + \frac{1}{r+1}\right) + H_{r+s}^{(2)} - H_r^{(2)} \right), $$

which holds for $0 \le r \in \mathbb{C} \setminus \mathbb{Z}^-$, $0 \ne s \in \mathbb{C} \setminus \mathbb{Z}^-$. A special case of this identity is

$$ \sum_{n=1}^{\infty} \frac{H_n H_{n+1}}{n(n+1)} = \zeta(2) + 2\zeta(3), $$

where $\zeta(2) = \pi^2/6$. The second example deals with two series involving the product $H_n O_n$. In general, these series exhibit a nontrivial structure, in the sense that they typically cannot be expressed in terms of elementary functions, but rather involve combinations of dilogarithms. Two such evaluations are

$$ \sum_{n=1}^{\infty} (-1)^n \binom{2n}{n} \frac{H_n O_n}{8^n} = \sqrt{\frac{2}{3}} \left( \mathrm{Li}_2\left(\sqrt{\frac{2}{3}}\right) - \mathrm{Li}_2\left(-\sqrt{\frac{2}{3}}\right) - \frac{\pi^2}{4} - \frac{1}{2} \ln\left(\frac{2}{3}\right) \ln 12 \right) $$
and

$$ \begin{aligned} \sum_{n=1}^{\infty} (-1)^n \frac{C_n H_n O_n}{4^n} = & \ 2 \left( \mathrm{Li}_2(\sqrt{2}) - \mathrm{Li}_2(-\sqrt{2}) \right) + 2\sqrt{2} \left( \mathrm{Li}_2\left(\frac{1}{\sqrt{2}}\right) - \mathrm{Li}_2\left(-\frac{1}{\sqrt{2}}\right) \right) \\ & + 2(2 + 2\sqrt{2} - \ln 2) \ln(\sqrt{2} + 1) - \frac{\pi^2}{2}(1 + \sqrt{2}) \\ & + (3\sqrt{2}\ln 2 - 6\sqrt{2} - 4) \ln 2 + i\pi \ln 2. \end{aligned} $$

The latter result is a symmetric expression in dilogarithms and an equivalent form is

$$ \begin{aligned} \sum_{n=1}^{\infty} (-1)^{n-1} \frac{C_n H_n O_n}{4^n} = & \ 2\sqrt{2} \left( (\sqrt{2}-1) \mathrm{Li}_2\left(\frac{\sqrt{2}}{2}\right) + \mathrm{Li}_2\left(-\frac{\sqrt{2}}{2}\right) \right) - 2(2+2\sqrt{2}-\ln 2) \ln(\sqrt{2}+1) \\ & + \frac{6\sqrt{2}-7}{12} \pi^2 - \frac{6\sqrt{2}-1}{2} \ln^2 2 + 2(3\sqrt{2}+2) \ln 2. \end{aligned} $$

However, we succeed in evaluating certain series involving the golden ratio $\alpha = (1 + \sqrt{5})/2$ in closed form. For instance, in Corollary 5.7 below we show that

$$ \sum_{n=1}^{\infty} \binom{2n}{n} \frac{H_n O_n}{4^n \alpha^n} = \frac{\pi^2}{12} \alpha + \frac{1}{2}(5 \ln \alpha - 4 \ln 2) \alpha \ln \alpha. $$

The next lemma will be used frequently in this paper.

**Lemma 1.1.** Let $(a_n)_{n \ge 1}$ be a sequence with generating function $f(x) = \sum_{n=1}^{\infty} a_n x^n$. Then we have

$$ \sum_{n=1}^{\infty} a_n H_n x^n = \int_0^1 \frac{f(x) - f(xt)}{1-t} dt \qquad (1.12) $$

and

$$ \sum_{n=1}^{\infty} a_n O_n x^n = \int_0^1 \frac{f(x) - f(xt^2)}{1-t^2} dt. \qquad (1.13) $$

*Proof.* The results follow directly from the representations

$$ H_n = \int_0^1 \frac{1-x^n}{1-x} dx \quad \text{and} \quad O_n = \int_0^1 \frac{1-x^{2n}}{1-x^2} dx. $$

As a simple application of Lemma 1.1, we obtain the generating function of the harmonic numbers. By choosing $a_n = 1$ for all $n$ and employing the geometric series $f(x) = x/(1-x)$, where $|x| < 1$, we have

$$ \sum_{n=1}^{\infty} H_n x^n = \int_0^1 \frac{\frac{x}{1-x} - \frac{xt}{1-xt}}{1-t} dt = \frac{x}{1-x} \int_0^1 \frac{1}{1-xt} dt = -\frac{\ln(1-x)}{1-x}, \quad |x| < 1. \qquad (1.14) $$

## 2. Series associated with known generating functions

In this section, we derive some interesting series identities from known generating functions, hereby focusing on the generating functions (1.2), (1.14), (1.10), and (1.11).
## 2.A. Series associated with generating function (1.2)

First, we state a lemma.

**Lemma 2.1.** For $u, v \in \mathbb{C} \setminus \mathbb{Z}^{-}$, we have

$$ \int_0^1 x^u (1-x)^v dx = \frac{1}{\binom{u+v+1}{u+1} (u+1)}, \qquad (2.15) $$

$$ \int_0^1 x^u (1-x)^v \ln x \, dx = - \frac{H_{u+v+1} - H_u}{\binom{u+v+1}{u+1} (u+1)}, \qquad (2.16) $$

$$ \int_0^1 x^u (1-x)^v \ln^2 x \, dx = \frac{(H_{u+v-1} - H_u)^2 + H_{u+v+1}^{(2)} - H_u^{(2)}}{\binom{u+v+1}{v+1} (v+1)}, \qquad (2.17) $$

and, in particular,

$$ \int_0^1 x^{u-1} \ln(1-x) \, dx = - \frac{H_u}{u}, \quad 0 \neq u \in \mathbb{C} \setminus \mathbb{Z}^{-}. \qquad (2.18) $$

*Proof.* The identity (2.15) is the well-known Beta function integral, while (2.16) is obtained by differentiating with respect to $u$. Identity (2.17) follows from differentiating (2.16) with respect to $u$. These identities are also listed in Gradshteyn and Ryzhik's book [GrRy07] and other tables.

**Theorem 2.2.** For $s - 1/2 \notin \mathbb{Z}^{-}$, $r \neq 0$ and $r + s - 1/2 \notin \mathbb{Z}^{-}$, we have

$$ \sum_{n=0}^{\infty} \frac{\binom{2n}{n} H_{n+r+s} - H_s}{2^{2n} \binom{n+r+s}{s+1}} = \frac{2(s+1)}{2s+1} \frac{H_{r+s-1/2} - H_{s-1/2}}{\binom{r+s-1/2}{r-1}}. \qquad (2.19) $$

In particular,

$$ \sum_{n=0}^{\infty} \frac{\binom{2n}{n} H_{n+r}}{2^{2n} n+r} = \frac{2^{2r+1} O_r}{\binom{2r}{r} r}, \quad r \neq 0, \qquad (2.20) $$

$$ \sum_{n=0}^{\infty} \frac{C_n H_{n+1}}{2^{2n}} = 4. \qquad (2.21) $$

*Proof.* Write $(1-x)/4$ for $x$ into the first identity of (1.2) and arrange to obtain

$$ \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n}} (1-x)^{n+r-1} x^s \ln x = (1-x)^{r-1} x^{s-1/2} \ln x. $$

Integrating from 0 to 1 using Lemma 2.1 results in the main identity (2.19). Identity (2.20) follows from identity (2.19) by setting $s = 0$ in conjunction with relation (1.7).

**Corollary 2.3.** We have

$$ \sum_{n=1}^{\infty} \frac{\binom{2n}{n} H_n}{2^{2n} n} = \frac{\pi^2}{3}. \qquad (2.22) $$

*Proof.* Write (2.20) as

$$ \sum_{n=1}^{\infty} \frac{\binom{2n}{n} H_{n+r}}{2^{2n} n+r} = \frac{2^{2r+1} O_r}{\binom{2r}{r} r} - \frac{H_r}{r}, \quad r \neq 0. $$

Thus,

$$ \sum_{n=1}^{\infty} \frac{\binom{2n}{n} H_n}{2^{2n} n} = 2 \lim_{r \to 0} \frac{O_r}{r} - \lim_{r \to 0} \frac{H_r}{r}. \qquad (2.23) $$
Now, using L'Hôpital's rule and (1.6), we obtain

$$ 
\begin{aligned}
\lim_{r \to 0} \frac{O_r}{r} &= \lim_{r \to 0} \frac{H_{2r} - \frac{1}{2}H_r}{r} = \lim_{r \to 0} \frac{2(\zeta(2) - H_{2r}^{(2)}) - \frac{1}{2}(\zeta(2) - H_r^{(2)})}{1} \\
&= 2\zeta(2) - \frac{1}{2}\zeta(2) = \frac{3}{2}\zeta(2)
\end{aligned}
\quad (2.24) $$

and

$$ \lim_{r \to 0} \frac{H_r}{r} = \lim_{r \to 0} \frac{\zeta(2) - H_r^{(2)}}{1} = \zeta(2). \quad (2.25) $$

Plug (2.24) and (2.25) in (2.23) to obtain (2.22).

**Corollary 2.4.** We have

$$ \sum_{n=0}^{\infty} \frac{C_n H_n}{2^{2n}} = 4 \ln 2, \quad \sum_{n=0}^{\infty} \frac{n}{n+1} \frac{C_n H_{n+1}}{2^{2n}} = \frac{2\pi^2}{3} - 4. $$

*Proof.* Working with (2.21) we can calculate

$$ \sum_{n=0}^{\infty} \frac{C_n H_{n+1}}{2^{2n}} = \sum_{n=0}^{\infty} \frac{C_n H_n}{2^{2n}} + \sum_{n=0}^{\infty} \frac{C_n}{2^{2n}(n+1)} = \sum_{n=0}^{\infty} \frac{C_n H_n}{2^{2n}} + 4 - 4 \ln 2, $$

where we used the fact that

$$ \sum_{n=0}^{\infty} \frac{C_n}{2^{2n}(n+1)} = 4 - 4 \ln 2. $$

This evaluation can be found in [AdFr25], for instance. The second identity is a combination of (2.21) with

$$ \sum_{n=0}^{\infty} \frac{C_n H_{n+1}}{2^{2n}(n+1)} = 8 - \frac{2\pi^2}{3}, $$

which is a special case of a more general result also derived in [AdFr25].

**Remark 2.5.** Using (1.7) in conjunction with $\binom{n-1/2}{m} = 2^{-2m}\binom{2n}{n}\binom{n}{m}\binom{2(n-m)}{n-m}^{-1}$, we can restate identity (2.19) in the form

$$ \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n}} \frac{H_{n+r+s} - H_s}{\binom{n+r+s}{s+1}} = \frac{s+1}{r} 2^{2r+1} \binom{2s}{s} \frac{O_{r+s} - O_s}{\binom{r+s}{r} \binom{2(r+s)}{r+s}}. $$

**Theorem 2.6.** We have

$$ \sum_{n=0}^{\infty} \frac{O_{n+1}}{(n+1)(2n+1)} = \frac{\pi^2}{6}, \quad (2.26) $$

$$ \sum_{n=0}^{\infty} \frac{C_n O_{n+3}}{2^{2n}(2n+5)} = \frac{8}{9} + \frac{\pi}{32} - \frac{\pi \ln 2}{4}, $$

$$ \sum_{n=0}^{\infty} \frac{C_n H_{n+r+1}}{2^{2n+1}(n+r+1)} = \frac{H_r}{r} - \frac{2^{2r}(O_{r+1} - 1)}{\binom{2(r-1)}{r-1}(2r-1)(2r+1)}, \quad r \neq 0, r \neq \pm \frac{1}{2}, $$

and more generally, for $0 \neq s, r \in \mathbb{C} \setminus \mathbb{Z}^{-}$, $r + s - 1/2 \notin \mathbb{Z}^{-}$,

$$ \sum_{n=0}^{\infty} \frac{C_n H_{n+r+s} - H_{s-1}}{2^{2n+1} \binom{n+r+s}{s}} = \frac{H_{r+s-1} - H_{s-1}}{\binom{r+s-1}{s}} - \frac{2s}{2s+1} \frac{H_{r+s-1/2} - H_{s-1/2}}{\binom{r+s-1/2}{r-1}}. \quad (2.27) $$
Proof. Write $(1-x)/4$ for $x$ in the second identity in (1.2) to obtain

$$ \sum_{n=0}^{\infty} \frac{C_n(1-x)^{n+1}}{2^{2n+1}} = 1 - x^{1/2}, $$

which can now be arranged as

$$ \sum_{n=0}^{\infty} \frac{C_n(1-x)^{n+r} x^{s-1}}{2^{2n+1}} \ln x = (1-x)^{r-1} x^{s-1} \ln x - (1-x)^{r-1} x^{s-1/2} \ln x, $$

which now produces (2.27) upon integration from 0 to 1, using Lemma 2.1.

**Remark 2.7.** Identity (2.26) is also recorded as identity (8) and generalized in a different direction in [AdFr25].

## 2.B. Series associated with generating function (1.14)

Next, we derive series associated with the generating function of harmonic numbers (1.14).

**Theorem 2.8.** We have

$$ \sum_{n=1}^{\infty} \frac{H_n}{(n+s)(n+s+1)} = \frac{H_s}{s}, \quad s \neq 0, \qquad (2.28) $$

$$ \sum_{n=1}^{\infty} \frac{H_n}{n(n+1)} = \frac{\pi^2}{6}, $$

$$ \sum_{n=1}^{\infty} \frac{2^{2n} H_n}{(2n + 2s - 1)(2n + 2s + 1)(2n + 2s + 3) \binom{2(n+s-1)}{n+s-1}} = \frac{4}{3s(2s+1)} \frac{O_{s+1} - 1}{\binom{2s}{s}}, $$

and more generally, $0 \le r \in \mathbb{C} \setminus \mathbb{Z}^-$, $0 \ne s \in \mathbb{C} \setminus \mathbb{Z}^-$,

$$ \sum_{n=1}^{\infty} \frac{H_n}{\binom{n+r+s+1}{r+2}} = \frac{r+2}{s} \frac{H_{r+s} - H_r}{\binom{r+s}{r}}. \qquad (2.29) $$

Proof. Write $1-x$ for $x$ in the generating function (1.14) and arrange as

$$ \sum_{n=1}^{\infty} H_n x^{r+1} (1-x)^{n+s-1} = -x^r (1-x)^{s-1} \ln x. $$

Now integrate from 0 to 1, making use of Lemma 2.1; this gives (2.29). The other identities are special cases at $r = s = 0$ and $r = 1/2$.

**Remark 2.9.** The identity (2.28) looks very interesting but it is just a restatement of the integral given in (2.18). This integral is also stated by Xu [Xu17, Eq. (2.11)], with the restriction $u > 0$. The equivalence of the forms follows from a simple calculation using (1.14):

$$ \frac{H_u}{u} = - \int_0^1 x^{u-1} \ln(1-x) dx = \int_0^1 x^{u-1} (1-x) \left( -\frac{\ln(1-x)}{1-x} \right) dx \\ = \sum_{n=0}^{\infty} H_n \left( \int_0^1 x^{n+u-1} dx - \int_0^1 x^{n+u} dx \right). $$

Indeed, in Theorem 2.2, Xu [Xu17] offers a formula for shifted harmonic series of the form

$$ \sum_{n=1}^{\infty} \frac{H_{n+u}}{(n+v)(n+w)} $$

with parameters $u, v$, and $w$.
**Theorem 2.10.** For $r \in \mathbb{C} \setminus \mathbb{Z}^{-}$ and $m \in \mathbb{Z}^{+}$,

$$ \sum_{n=1}^{\infty} \frac{H_n}{(n+r+1)^{m+1}} = H_r(\zeta(m+1) - H_r^{(m+1)}) + \frac{m+1}{2}(\zeta(m+2) - H_r^{(m+2)}) \\ - \frac{1}{2}\sum_{n=1}^{m-1}(\zeta(n+1) - H_r^{(n+1)})(\zeta(m-n+1) - H_r^{(m-n+1)}). \qquad (2.30) $$

In particular,

$$ \sum_{n=1}^{\infty} \frac{H_n}{(n+r+1)^2} = H_r(\zeta(2) - H_r^{(2)}) + \zeta(3) - H_r^{(3)}. $$

*Proof.* Sum (2.28) over $s$ from 1 to $r$ to obtain

$$ \sum_{n=1}^{\infty} H_n \left( \frac{1}{n+1} - \frac{1}{n+r+1} \right) = \frac{1}{2} (H_r^2 + H_r^{(2)}), \qquad (2.31) $$

which holds for all complex numbers $r$ that is not a negative integer. Identity (2.30) then follows upon differentiating (2.31) $m$ times with respect to $r$, using Leibnitz rule and the following facts:

$$ \frac{d^m}{dr^m} H_r = (-1)^{m-1} m! (\zeta(m+1) - H_r^{(m+1)}) $$

and

$$ \frac{d^m}{dr^m} H_r^{(2)} = (-1)^{m-1} (m+1)! (\zeta(m+2) - H_r^{(m+2)}). $$

**Remark 2.11.** Identity (2.30) is equivalent to a main result of Sofo and Cvijović [SoCv12]. While their identity is expressed in terms of the Hurwitz zeta function, our formulation is given directly in terms of the Riemann zeta function.

**Theorem 2.12.** We have

$$ \begin{aligned} \sum_{n=1}^{\infty} \frac{H_n H_{n+1}}{n(n+1)} &= \zeta(2) + 2\zeta(3), \\ \sum_{n=1}^{\infty} \frac{H_n O_n}{(2n-3)(2n-1)} &= -\frac{1}{6} + \frac{\pi^2}{36} + \frac{1}{2}\ln 2, \\ \sum_{n=1}^{\infty} \frac{H_n H_{n+s+1}}{(n+s)(n+s+1)} &= \frac{1}{s}(H_s + H_s^2 + H_s^{(2)}), \quad s \neq 0, \end{aligned} $$

and more generally, for $0 \le r \in \mathbb{C} \setminus \mathbb{Z}^{-}$, $0 \ne s \in \mathbb{C} \setminus \mathbb{Z}^{-}$,

$$ \sum_{n=1}^{\infty} \frac{H_n H_{n+r+s+1}}{\binom{n+r+s+1}{r+2}} = \frac{r+2}{s \binom{r+s}{r}} \left( (H_{r+s} - H_r) \left(H_{r+s} + \frac{1}{r+1}\right) + H_{r+s}^{(2)} - H_r^{(2)} \right). $$

*Proof.* Using (1.14), write

$$ \sum_{n=1}^{\infty} H_n (1-x)^{n+s-1} x^{r+1} \ln x = -(1-x)^{s-1} x^r \ln^2 x $$

and integrate from 0 to 1 using Lemma 2.1.
## 2.C. Series associated with generating function (1.10)

Next, we establish some identities associated with the generating function (1.10).

**Theorem 2.13.** We have

$$ \sum_{n=1}^{\infty} \frac{\binom{2n}{n} \binom{2(n-1)}{n-1} O_n}{2^{4n} n} = \frac{\ln 2}{\pi}, \quad \sum_{n=1}^{\infty} \frac{\binom{2n}{n}^2 O_n}{2^{4n}(n+1)} = \frac{4(1 - \ln 2)}{\pi}, $$

and more generally, if r is a non-negative integer, then

$$ \sum_{n=1}^{\infty} \frac{\binom{2n}{n} \binom{2(n+r-1)}{n+r-1} O_n}{2^{4n}(n+r)} = \frac{2^{2r} O_r - \ln 2}{2r-1 \pi} \quad (2.32) $$

*Proof.* Write $x/4$ for $x$ in (1.10) and arrange as

$$ \sum_{n=1}^{\infty} \binom{2n}{n} \frac{O_n}{2^{2n}} x^{n+r-1} (1-x)^{1/2} = -\frac{1}{2} x^{r-1} \ln(1-x). $$

Integrate both sides from 0 to 1 using Lemma 2.1 and substitute $r - 1/2$ for $r$ to obtain

$$ \sum_{n=1}^{\infty} \frac{\binom{2n}{n} O_n}{2^{2n} \binom{n+r}{3/2}} = \frac{3H_{r-1/2}}{2(2r-1)} $$

and hence deduce (2.32), since

$$ \binom{q}{3/2} = \frac{q!}{\left(\frac{3}{2}\right)!\left(q-1-\frac{1}{2}\right)!} = \frac{q!}{\left(\frac{3\sqrt{\pi}}{4}\right)!} \frac{2^{2(q-1)}}{\sqrt{\pi} \binom{2(q-1)}{q-1}} \left(q-1\right)! = \frac{2^{2q} q}{3\pi \binom{2(q-1)}{q-1}} $$

and using the identity $H_{k-1/2} = 2O_k - 2\ln 2$.

**Theorem 2.14.** We have

$$ \sum_{n=1}^{\infty} \frac{\binom{2n}{n} O_n}{2^{2n} n} = \sum_{n=1}^{\infty} \frac{2^{2n} H_n}{n(n+1) \binom{2(n+1)}{n+1}}, \\ \sum_{n=1}^{\infty} \frac{C_n O_n}{2^{2n}} = \sum_{n=1}^{\infty} \frac{2^{2(n+1)} H_n}{(n+1)(n+2) \binom{2(n+2)}{n+2}}, $$

and more generally, if r is a non-negative integer, then

$$ \sum_{n=1}^{\infty} \frac{\binom{2n}{n} O_n}{2^{2n}(n+r)} = \sum_{n=1}^{\infty} \frac{2^{2(n+r)} H_n}{(n+r)(n+r+1) \binom{2(n+r+1)}{n+r+1}} \quad (2.33) $$

*Proof.* We have

$$ \begin{aligned} \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}} O_n x^{n+r-1} &= -\frac{x^{r-1} \ln(1-x)}{2 \sqrt{1-x}} \\ &= -x^{r-1} \frac{\sqrt{1-x}}{2} \sum_{n=1}^{\infty} x^n H_n = -\frac{1}{2} \sum_{n=1}^{\infty} x^{n+r-1} (1-x)^{1/2} H_n; \end{aligned} $$
from which, integrating from 0 to 1, using Lemma 2.1, we get

$$ \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}} \frac{O_n}{n+r} = \frac{1}{3} \sum_{n=1}^{\infty} \frac{H_n}{\binom{n+r+1/2}{3/2}} $$

and hence (2.33), since

$$ \binom{q+1/2}{3/2} = \binom{q+1/2}{q-1} = \frac{\frac{\sqrt{\pi}}{2^{2q+2}} \binom{2q+2}{q+1} (q+1)!}{\frac{3\sqrt{\pi}}{4} (q-1)!} = \frac{q(q+1)}{2^{2q3}} \binom{2q+2}{q+1} $$

$$ \text{and } (q+1/2)! = \sqrt{\pi} 2^{-2q-2} \binom{2q+2}{q+1} (q+1)! $$

**Remark 2.15.** Combining the previous results, we obtain

$$ \sum_{n=1}^{\infty} \frac{C_n O_n}{2^{2n} n} = \frac{3}{2} \sum_{n=1}^{\infty} \frac{2^{2n}}{n(n+1)(2n+1)(2n+3)} \frac{H_n}{C_n} $$

Next, we present selected series evaluations involving products of Fibonacci (Lucas) numbers and odd harmonic numbers. Recall that the Fibonacci and Lucas numbers both satisfy the recurrence $u_n = u_{n-1} + u_{n-2}$, $n \ge 2$, but with respective initial conditions $F_0 = 0$, $F_1 = 1$ and $L_0 = 2$, $L_1 = 1$. For negative subscripts, we have $F_{-n} = (-1)^{n-1}F_n$ and $L_{-n} = (-1)^n L_n$.

The Binet formulas for these numbers state that

$$ F_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}, \quad L_n = \alpha^n + \beta^n, \quad n \in \mathbb{Z}, \qquad (2.34) $$

$$ \text{where } \alpha = (1 + \sqrt{5})/2 \text{ and } \beta = -1/\alpha. $$

**Theorem 2.16.** If r is an even integer and s any integer, then

$$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \binom{2n}{n} O_n F_{rn+s}}{2^{2n-1} L_r^n} = \frac{1}{\sqrt{5}} \left( \frac{\alpha^s \ln(1 + \alpha^r / L_r)}{\sqrt{1 + \alpha^r / L_r}} - \frac{\beta^s \ln(1 + \beta^r / L_r)}{\sqrt{1 + \beta^r / L_r}} \right), \\ \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \binom{2n}{n} O_n L_{rn+s}}{2^{2n-1} L_r^n} = \frac{\alpha^s \ln(1 + \alpha^r / L_r)}{\sqrt{1 + \alpha^r / L_r}} + \frac{\beta^s \ln(1 + \beta^r / L_r)}{\sqrt{1 + \beta^r / L_r}} $$

*Proof.* Insert $x = -\alpha^r/L_r$ and $x = -\beta^r/L_r$, in turn, into (1.10), then multiply through by $\alpha^s$ (or $\beta^s$) and combine the resulting expressions according to the Binet formulas (2.34).

**Example 2.17.** Choosing $s = 0$ and $r = 0$, $r = 1$, $r = 2$ in Theorem 2.16 gives:

$$ \begin{aligned} & \sum_{n=1}^{\infty} (-1)^{n-1} \frac{O_n}{8^n} = \frac{1}{\sqrt{6}} \ln\left(\frac{3}{2}\right), \\ & \sum_{n=1}^{\infty} (-1)^{n-1} \frac{O_n L_{2n}}{12^n} = \frac{\sqrt{57}}{38} \left( \sqrt{5-\alpha} \ln\left(\frac{\alpha+4}{3}\right) + \sqrt{4+\alpha} \ln\left(\frac{5-\alpha}{3}\right) \right), \\ & \sum_{n=1}^{\infty} (-1)^{n-1} \frac{O_n F_{2n}}{12^n} = \frac{\sqrt{285}}{190} \left( \sqrt{5-\alpha} \ln\left(\frac{\alpha+4}{3}\right) - \sqrt{4+\alpha} \ln\left(\frac{5-\alpha}{3}\right) \right), \\ & \sum_{n=1}^{\infty} (-1)^{n-1} \frac{O_n L_{4n}}{28^n} = \frac{\sqrt{231}}{66} \left( \sqrt{4-\alpha} \ln\left(\frac{3\alpha+9}{7}\right) + \sqrt{3+\alpha} \ln\left(\frac{12-3\alpha}{7}\right) \right), \\ & \sum_{n=1}^{\infty} (-1)^{n-1} \frac{O_n F_{4n}}{28^n} = \frac{\sqrt{1155}}{330} \left( \sqrt{4-\alpha} \ln\left(\frac{3\alpha+9}{7}\right) - \sqrt{3+\alpha} \ln\left(\frac{12-3\alpha}{7}\right) \right). \end{aligned} $$
## 2.D. Series associated with generating function (11)

We now derive identities associated with generating function (1.11).

**Theorem 2.18.** We have

$$ \sum_{n=1}^{\infty} \frac{C_n O_n H_{n+1}}{2^{2n}(n+1)} = 24 - 14\zeta(3) - \frac{2\pi^2}{3}, \qquad (2.35) $$

$$ \sum_{n=1}^{\infty} \frac{C_n O_n H_{n+2}}{2^{2n}(n+2)} = \frac{14}{27}, \qquad (2.36) $$

and more generally, for $s - 1/2 \notin \mathbb{Z}^{-}$, $0 \neq s \in \mathbb{C} \setminus \mathbb{Z}^{-}$, $r + s + 1/2 \notin \mathbb{Z}^{-}$,

$$ \sum_{n=1}^{\infty} \frac{C_n O_n}{2^{2n}} \frac{H_{n+r+s+1} - H_{s-1}}{\binom{n+r+s+1}{s}} = \frac{2(H_{r+s} - H_{s-1})}{\binom{r+s}{s}} - \frac{4s}{2s+1} \frac{H_{r+s+1/2} - H_{s-1/2}}{\binom{r+s+1/2}{r}} \\ - \frac{2s}{2s+1} \frac{(H_{r+s+1/2} - H_{s-1/2})^2 + H_{r+s+1/2}^{(2)} - H_{s-1/2}^{(2)}}{\binom{r+s+1/2}{r}}. \qquad (2.37) $$

*Proof.* Write $(1-x)/4$ for $x$ in (1.11) and rearrange to obtain

$$ \sum_{n=1}^{\infty} \frac{C_n O_n}{2^{2n}} (1-x)^{n+r+1} x^{s-1} = 2(1-x)^r x^{s-1} - 2(1-x)^r x^{s-1/2} + (1-x)^r x^{s-1/2} \ln x. $$

Now integrate both sides from 0 to 1 using Lemma 2.1. We also note that identity (2.35) was obtained by setting $s = 1$ in (2.37) and taking the limit at $r = -1$.

## 3. Series associated with the generating function of generalized harmonic numbers

In [Xu17, Lemma 1.2], Xu established the following result.

**Lemma 3.1.** For $0 \neq s \in \mathbb{C} \setminus \mathbb{Z}^{-}$ and $p \in \mathbb{Z}^{+}$,

$$ \int_{0}^{1} x^{s-1} \mathrm{Li}_{p}(x) \, dx = \sum_{n=1}^{p-1} \frac{(-1)^{n-1}}{s^n} \zeta(p+1-n) - (-1)^p \frac{H_s}{s^p}. \qquad (3.38) $$

**Theorem 3.2.** For $0 \neq s \in \mathbb{C} \setminus \mathbb{Z}^{-}$ and $p \in \mathbb{Z}^{+}$,

$$ \sum_{n=1}^{\infty} \frac{H_n^{(p)}}{(n+s)(n+s+1)} = \sum_{n=1}^{p-1} \frac{(-1)^{n-1}}{s^n} \zeta(p+1-n) - (-1)^p \frac{H_s}{s^p}. \qquad (3.39) $$

*Proof.* The generating function of the generalized harmonic numbers is given by

$$ \sum_{n=1}^{\infty} H_n^{(p)} x^n = \frac{\mathrm{Li}_p(x)}{1-x}, $$

which allows the following rearrangement:

$$ \sum_{n=1}^{\infty} H_n^{(p)} (1-x) x^{n+s-1} = x^{s-1} \mathrm{Li}_p(x), $$

from which (3.39) follows by integration, using (3.38).
**Remark 3.3.** Identity (3.39) generalizes identity (2.28).

**Theorem 3.4.** If $p$ is a positive integer and $m$ and $r$ are non-negative integers, then

$$ \sum_{n=1}^{\infty} \frac{H_{n+r}^{(m+1)} - H_{n}^{(m+1)}}{n^p} = \sum_{n=1}^{p-1} (-1)^{n-1} \binom{m+1-n}{m} \zeta(p+1-n) H_r^{(m+n)} \\ + (-1)^p \sum_{n=1}^{m} \binom{p+m-n-1}{m-n} \zeta(n+1) H_r^{(m+p-n)} \\ - (-1)^p \sum_{n=0}^{m} \sum_{j=1}^{r} \binom{p+m-n-1}{m-n} \frac{H_j^{(n+1)}}{j^{p+m-n}}. \qquad (3.40) $$

In particular,

$$ \sum_{n=1}^{\infty} \frac{H_{n+r}}{n^2} = 2\zeta(3) + \zeta(2)H_r - \sum_{n=1}^{r} \frac{H_n}{n^2}, $$

where we used $\sum_{n=1}^{\infty} \frac{H_n}{n^2} = 2\zeta(3)$.

*Proof.* We begin with the polylogarithm series

$$ \sum_{n=1}^{\infty} \frac{x^n}{n^p} = \mathrm{Li}_p(x), $$

which allows us to write

$$ \sum_{n=1}^{\infty} \frac{x^{n+s-1}}{n^p} = x^{s-1} \mathrm{Li}_p(x), $$

from which, integrating from 0 to 1, we get, by (3.38),

$$ \sum_{n=1}^{\infty} \frac{1}{n^p (n+s)} = \sum_{n=1}^{p-1} \frac{(-1)^{n-1}}{s^n} \zeta(p+1-n) - (-1)^p \frac{H_s}{s^p}. \qquad (3.41) $$

Differentiating (3.41) $m$ times with respect to $s$ and simplifying, we find

$$ \sum_{n=1}^{\infty} \frac{1}{n^p (n+s)^{m+1}} = \sum_{n=1}^{p-1} (-1)^{n-1} \binom{m+1-n}{m} \frac{\zeta(p+1-n)}{s^{m+n}} \\ + (-1)^p \sum_{n=1}^{m} \binom{p+m-n-1}{m-n} \frac{\zeta(n+1)}{s^{p+m-n}} - (-1)^p \sum_{n=0}^{m} \binom{p+m-n-1}{m-n} \frac{H_s^{(n+1)}}{s^{p+m-n}}. \qquad (3.42) $$

Summing (3.42) over $s$ from 1 to $r$ gives (3.40).

**Corollary 3.5.** If $p \ge 2$ is an integer and $r$ is a non-negative integer, then

$$ \sum_{n=1}^{\infty} \frac{H_{n+r}}{n^p} = \frac{p+2}{2} \zeta(p+1) - \frac{1}{2} \sum_{n=2}^{p-1} \zeta(n) \zeta(p+1-n) \\ + \sum_{n=1}^{p-1} (-1)^{n-1} \zeta(p+1-n) H_r^{(n)} - (-1)^p \sum_{n=1}^{r} \frac{H_n}{n^p}. $$
Proof. Set $m = 0$ in (3.40) and use the standard Euler sum

$$ \sum_{n=1}^{\infty} \frac{H_n}{n^p} = \frac{p+2}{2} \zeta(p+1) - \frac{1}{2} \sum_{n=2}^{p-1} \zeta(n) \zeta(p+1-n). $$

**Remark 3.6.** An explicit evaluation of each of the following sums is possible: $\sum_{n=1}^{\infty} \frac{H_{n+r}^{(p)}}{n^p}$, $\sum_{n=1}^{\infty} \frac{H_{n+r}^{(m+1)}}{n^p}$ (when $m + p$ is even integer), $\sum_{n=1}^{\infty} \frac{H_{n+r}^{(2)}}{n^4}$, and $\sum_{n=1}^{\infty} \frac{H_{n+r}^{(4)}}{n^2}$. This is due to the well-known fact that the sum $\sum_{n=1}^{\infty} \frac{H_n^{(q)}}{n^p}$ can be expressed in terms of zeta values in the special cases $q = 1, p = q$ or $p + q$ odd or $p + q$ even, but with $(p, q)$ being restricted to $(2, 4)$ or $(4, 2)$.

## 4. Series associated with the dilogarithm

We require the integration formulas stated in the next lemma.

**Lemma 4.1.** For $\Re(s) > 0$, we have

$$ \int_0^1 x^{s-1} \mathrm{Li}_2(x) dx = \frac{\pi^2}{6s} - \frac{H_s}{s^2}, \qquad (4.43) $$

$$ \int_0^1 (1-x)^{s-1} \mathrm{Li}_2(x) dx = \frac{\pi^2}{6s} - \frac{H_s^{(2)}}{s}, \qquad (4.44) $$

$$ \int_0^1 x^{s-1} \mathrm{Li}_2(x) \ln x \, dx = -\frac{\pi^2}{3s^2} + \frac{H_s^{(2)}}{s^2} + \frac{2H_s}{s^3}, \qquad (4.45) $$

and

$$ \int_0^1 (1-x)^{s-1} \mathrm{Li}_2(x) \ln(1-x) dx = -\frac{\pi^2}{6s^2} - \frac{2\zeta(3)}{s} + \frac{2H_s^3}{s} + \frac{H_s^{(2)}}{s^2}. \qquad (4.46) $$

Proof. Identity (4.43) is recorded in Lewin [Lew81, p. 308] (also $p = 2$ in (3.38)).
Multiply through the dilogarithm functional equation

$$ \mathrm{Li}_2(x) + \mathrm{Li}_2(1-x) = \frac{\pi^2}{6} - \ln x \ln(1-x), $$

by $x^{u-1}$ and integrate from 0 to 1, using (4.43) and the known integral (e.g., Xu [Xu17, Eq. (2.13)]):

$$ \int_0^1 x^{s-1} \ln x \ln(1-x) dx = -\frac{\pi^2}{6s} + \frac{H_s^{(2)}}{s} + \frac{H_s}{s^2}; $$

this gives (4.44). Identities (4.45) and (4.46) are obtained by differentiating (4.43) and (4.44) with respect to $u$.

**Theorem 4.2.** We have

$$ \sum_{n=1}^{\infty} \frac{H_{n+1}^{(2)}}{n(n+1)} = 3 - \frac{\pi^2}{6}. $$

Proof. Write $\sum_{n=1}^{\infty} \frac{(1-x)^n}{n}$ $\mathrm{Li}_2(x) = -\ln x \mathrm{Li}_2(x)$ and integrate.
**Theorem 4.3.** If $0 \neq s \in \mathbb{C} \setminus \mathbb{Z}^{-}$, then

$$ \sum_{n=1}^{\infty} \binom{n+s}{n+1}^{-1} \frac{H_{n+s} - H_{s-1}}{(n+1)n^2} = \frac{\zeta(2)}{s^2} + \frac{2\zeta(3)}{s} - \frac{2H_s^{(3)}}{s} - \frac{H_s^{(2)}}{s^2}. \quad (4.47) $$

In particular,

$$ \sum_{n=1}^{\infty} \frac{H_{n+1}}{n^2(n+1)} = \zeta(2) + 2\zeta(3) - 3. $$

*Proof.* The series $\sum_{n=1}^{\infty} \frac{x^n}{n^2} = \mathrm{Li}_2(x)$ allows us to write

$$ \sum_{n=1}^{\infty} \frac{(1-x)^{s-1} x^n}{n^2} \ln(1-x) = (1-x)^{s-1} \mathrm{Li}_2(x) \ln(1-x), $$

from which (4.47) follows upon term-wise integration from 0 to 1, using (2.16) and (4.46).

**Corollary 4.4.** We have

$$ \sum_{n=1}^{\infty} 2^{2n} \binom{2(n+1)}{n+1}^{-1} \frac{O_{n+1}}{n^2(n+1)} = \frac{3}{2}\zeta(2) + \frac{7}{2}\zeta(3) - 6. $$

*Proof.* Set $x = 1/2$ in (4.47) and use $\binom{n+1/2}{n+1} = 2^{-2(n+1)}\binom{2(n+1)}{n+1}$, along with the relations

$$ H_{n+1/2} - H_{-1/2} = 2O_{n+1}, \quad H_{1/2}^{(3)} = 8 - 6\zeta(3), \quad H_{1/2}^{(2)} = 4 - 2\zeta(2). $$

The next two results are derived from the following generating function of $H_n^2$, which is given by (see, for example, [Cof03]):

$$ \sum_{n=1}^{\infty} H_n^2 x^n = \frac{\ln^2(1-x)}{1-x} + \frac{\mathrm{Li}_2(x)}{1-x}. \quad (4.48) $$

**Theorem 4.5.** For $0 \neq s \in \mathbb{C} \setminus \mathbb{Z}^{-}$,

$$ \sum_{n=1}^{\infty} \frac{H_n^2}{(n+s)(n+s+1)} = \frac{H_s H_{s-1}}{s} + \frac{H_s^{(2)}}{s} + \frac{\zeta(2)}{s}. $$

In particular,

$$ \sum_{n=1}^{\infty} \frac{H_n^2}{(n+1)(n+2)} = \zeta(2) + 1. $$

*Proof.* Using (4.48), set up the following identity

$$ \sum_{n=1}^{\infty} H_n^2 x^{n+s-1} (1-x) = x^{s-1} \ln^2(1-x) + x^{s-1} \mathrm{Li}_2(x) \quad (4.49) $$

and integrate, using (2.17) and Lemma 4.1 to evaluate the resulting integrals.

**Theorem 4.6.** For $0 \neq s \in \mathbb{C} \setminus \mathbb{Z}^{-}$,

$$ \sum_{n=1}^{\infty} \frac{H_n^2}{\binom{n+s+1}{s+1}} = \frac{s+1}{s} \left( \frac{2}{s^2} + \zeta(2) - H_s^{(2)} \right). \quad (4.50) $$

*Proof.* Write

$$ \sum_{n=1}^{\infty} H_n^2 x^n (1-x)^s = (1-x)^{s-1} \ln^2(1-x) + (1-x)^{s-1} \mathrm{Li}_2(x), $$

and integrate.
## 5. A first new generating function

The goal of this section is to derive a new generating function involving central binomial coefficients, harmonic numbers, and odd harmonic numbers. Our first result in this section is a new proof of the generating function (1.10) (see [Ch16, Theorem 8] or [AFG25, Theorem 3]).

**Theorem 5.1.** For all $x \in [-1/4, 1/4)$, we have

$$ \sum_{n=1}^{\infty} \binom{2n}{n} O_n x^n = - \frac{\ln(1-4x)}{2\sqrt{1-4x}}. \qquad (5.51) $$

*Proof.* For the sequence $a_n = \binom{2n}{n}$, we use (1.2) to see that $f(x) = \frac{1-\sqrt{1-4x}}{\sqrt{1-4x}}$ and by series-integral correspondence (1.13), we have

$$ \sum_{n=1}^{\infty} \binom{2n}{n} O_n x^n = \int_0^1 \frac{1}{1-t^2} \frac{\sqrt{1-4xt^2} - \sqrt{1-4x}}{\sqrt{1-4x}\sqrt{1-4xt^2}} dt. $$

Next, as we have

$$ \int \frac{1}{1-t^2} \frac{\sqrt{1-4xt^2} - \sqrt{1-4x}}{\sqrt{1-4x}\sqrt{1-4xt^2}} dt = \frac{1}{\sqrt{1-4x}} \left( \arctanh t - \arctanh \left( \frac{\sqrt{1-4x}t}{\sqrt{1-4xt^2}} \right) \right) + c, $$

we obtain the identity

$$ \sum_{n=1}^{\infty} \binom{2n}{n} O_n x^n = \frac{1}{\sqrt{1-4x}} \lim_{t \to 1^{-}} \left( \arctanh t - \arctanh \left( \frac{\sqrt{1-4x}t}{\sqrt{1-4xt^2}} \right) \right), $$

where the limit is of the form $\infty - \infty$. Now, using the definition of the inverse hyperbolic tangent function, we get

$$ \begin{aligned} \lim_{t \to 1^{-}} \left( \arctanh t - \arctanh \left( \frac{\sqrt{1-4x}t}{\sqrt{1-4xt^2}} \right) \right) &= \frac{1}{2} \lim_{t \to 1^{-}} \left( \ln \left( 1 - \frac{\sqrt{1-4x}t}{\sqrt{1-4xt^2}} \right) - \ln(1-t) \right) \\ &= \frac{1}{2} \lim_{t \to 1^{-}} \left( \ln \left( \frac{\sqrt{1-4xt^2} - \sqrt{1-4x}t}{1-t} \right) - \ln(\sqrt{1-4xt^2}) \right) \\ &= \frac{1}{2} \left( \ln \left( \frac{1}{\sqrt{1-4x}} \right) - \ln(\sqrt{1-4x}) \right) = -\ln(\sqrt{1-4x}), \end{aligned} $$

where we have used the continuity of the logarithm and

$$ \lim_{t \to 1^{-}} \frac{\sqrt{1-4xt^2} - \sqrt{1-4x}t}{1-t} = \frac{1}{\sqrt{1-4x}}. $$

This completes the proof.

**Lemma 5.2.** For all $x \in [-1/4, 1/4)$,

$$ F(x) = \lim_{t \to 1^{-}} \left( \ln(1-4x) \ln(1-t) - \ln(1-4xt) \ln\left(\sqrt{1-4x} - \sqrt{1-4xt}\right) \right) \\ = \begin{cases} \ln(1-4x) \ln\left(-\frac{\sqrt{1-4x}}{2x}\right), & \text{if } x < 0; \\ 0, & \text{if } x = 0; \\ \ln(1-4x) \left(\ln\left(\frac{\sqrt{1-4x}}{2x}\right) - i\pi\right), & \text{if } x > 0, \end{cases} \qquad (5.52) $$

where $i = \sqrt{-1}$.
*Proof.* We find

$$ 
\begin{aligned}
\lim_{t \to 1} & \left( \ln(1-4x) \ln(1-t) - \ln(1-4xt) \ln(\sqrt{1-4x} - \sqrt{1-4xt}) \right) \\
& = \ln(1-4x) \lim_{t \to 1} \left( \ln(1-t) - \ln(\sqrt{1-4x} - \sqrt{1-4xt}) \right) \\
& = \ln(1-4x) \lim_{t \to 1} \ln \left( \frac{1-t}{\sqrt{1-4x} - \sqrt{1-4xt}} \right),
\end{aligned}
$$

and by applying L'Hopital's rule, we obtain the result

$$ F(x) = \ln(1-4x) \ln \left( -\frac{\sqrt{1-4x}}{2x} \right). $$

**Theorem 5.3.** For all $x \in [-1/4, 1/4)$, we have

$$ \sum_{n=1}^{\infty} \binom{2n}{n} H_n O_n x^n = \frac{1}{\sqrt{1-4x}} \left( \mathrm{Li}_2\left(\frac{1}{\sqrt{1-4x}}\right) - \mathrm{Li}_2\left(-\frac{1}{\sqrt{1-4x}}\right) - \frac{\pi^2}{4} + G(x) \right) \quad (5.53) $$

with

$$ G(x) = \frac{1}{2} \ln(1-4x) \begin{cases} \ln\left(-\frac{1-4x}{x}\right), & \text{if } x < 0; \\ \ln\left(\frac{1-4x}{x}\right) - i\pi, & \text{if } x > 0. \end{cases} \quad (5.54) $$

For $x = 0$ the series on the left-hand side of (5.53) is obviously zero.

*Proof.* We apply the previous theorem with $a_n = \binom{2n}{n}O_n$ and $f(x) = -\frac{\ln(1-4x)}{2\sqrt{1-4x}}$, and combine it with (1.12). This yields

$$ \sum_{n=1}^{\infty} \binom{2n}{n} H_n O_n x^n = \frac{I(x)}{2\sqrt{1-4x}}, $$

where

$$ I(x) = \int_{0}^{1} \frac{\sqrt{1-4x} \ln(1-4xt) - \sqrt{1-4xt} \ln(1-4x)}{(1-t)\sqrt{1-4xt}} dt. $$

Now, one checks that

$$ 
\begin{aligned}
& \int \frac{\sqrt{1-4x} \ln(1-4xt) - \sqrt{1-4xt} \ln(1-4x)}{(1-t)\sqrt{1-4xt}} dt \\
& = 2 \mathrm{Li}_2\left(-\frac{\sqrt{1-4xt}}{\sqrt{1-4x}}\right) - 2 \mathrm{Li}_2\left(\frac{\sqrt{1-4xt}}{\sqrt{1-4x}}\right) + \ln(1-4x) \ln(1-t) \\
& \quad + 2 \ln(1-4xt) \operatorname{arctanh}\left(\frac{\sqrt{1-4xt}}{\sqrt{1-4x}}\right) + c.
\end{aligned}
$$

This produces

$$ I(x) = 2 \mathrm{Li}_2\left(\frac{1}{\sqrt{1-4x}}\right) - 2 \mathrm{Li}_2\left(-\frac{1}{\sqrt{1-4x}}\right) + 2 \mathrm{Li}_2(-1) - 2 \mathrm{Li}_2(1) \\ 
+ \lim_{t \to 1^{-}} \left( \ln(1-4x) \ln(1-t) + 2 \ln(1-4xt) \operatorname{arctanh}\left(\frac{\sqrt{1-4xt}}{\sqrt{1-4x}}\right) \right). $$

To determine the limit we again use the definition of the arctanh function and simplify. The result is

$$ 
\begin{aligned}
& \lim_{t \to 1^{-}} \left( \ln(1-4x) \ln(1-t) + 2 \ln(1-4xt) \operatorname{arctanh}\left(\frac{\sqrt{1-4xt}}{\sqrt{1-4x}}\right) \right) \\
& = \ln(1-4x) \ln(2\sqrt{1-4x}) + F(x).
\end{aligned}
$$

The result follows from (5.52) and further simplification steps.
**Remark 5.4.** For all $x \in [-1, 0)$ (and analogously for $x \in (0, 1)$), identity (5.53) can also be written as

$$ \sum_{n=1}^{\infty} \binom{2n}{n} \frac{H_n O_n}{4^n} x^n = \frac{1}{\sqrt{1-x}} \left( \mathrm{Li}_2\left(\frac{1}{\sqrt{1-x}}\right) - \mathrm{Li}_2\left(-\frac{1}{\sqrt{1-x}}\right) \right. \\ \left. - \frac{\pi^2}{4} + \frac{1}{2} \ln(1-x) \ln\left(-\frac{4(1-x)}{x}\right) \right). \quad (5.55) $$

In addition, we mention that an equivalent form of (5.53) is

$$ \sum_{n=1}^{\infty} \binom{2n}{n} H_n O_n x^n = \frac{1}{2\sqrt{1-4x}} \times \left( \mathrm{Li}_2(1-4x) - 4 \mathrm{Li}_2(\sqrt{1-4x}) + \frac{\pi^2}{2} + \ln(1-4x) \left( \ln\left(\frac{1-4x}{|x|}\right) - \frac{1-\operatorname{sgn}x}{2} \pi i \right) \right), $$

so that

$$ \sum_{n=1}^{\infty} \binom{2n}{n} \frac{H_n O_n}{4^n} x^n = \frac{1}{2\sqrt{1-x}} \left( \mathrm{Li}_2(1-x) - 4 \mathrm{Li}_2(\sqrt{1-x}) + \frac{\pi^2}{2} \right. \\ \left. + \ln(1-x) \left( \ln\left(\frac{4(x-1)}{x}\right) - i\pi \right) \right). $$

**Remark 5.5.** To prove Theorem 5.3, an equivalent approach would be to start with $a_n = \binom{2n}{n} H_n$, use Boyadzhiev's expression (1.8) and apply (1.13). So, what we have shown above is the evaluation

$$ \int_0^1 \left( \frac{2}{\sqrt{1-4x}} \ln \left( \frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}} \right) - \frac{2}{\sqrt{1-4xt^2}} \ln \left( \frac{1+\sqrt{1-4xt^2}}{2\sqrt{1-4xt^2}} \right) \right) \frac{dt}{1-t^2} \\ = \frac{1}{\sqrt{1-4x}} \left( \mathrm{Li}_2\left(\frac{1}{\sqrt{1-4x}}\right) - \mathrm{Li}_2\left(-\frac{1}{\sqrt{1-4x}}\right) - \frac{\pi^2}{4} + G(x) \right), $$

where $G(x)$ is given in (5.54).

**Corollary 5.6.** We have

$$ \sum_{n=1}^{\infty} (-1)^n \binom{2n}{n} \frac{H_n O_n}{4^n} = \frac{\sqrt{2}}{8} \left( 8 \mathrm{Li}_2\left(\frac{1}{\sqrt{2}}\right) - \frac{7\pi^2}{6} + 7 \ln^2 2 \right), $$

$$ \sum_{n=1}^{\infty} \binom{2n}{n} \frac{H_n O_n}{8^n} = \sqrt{2} \left( -\mathrm{Li}_2\left(\frac{1}{\sqrt{2}}\right) - \mathrm{Li}_2(-\sqrt{2}) + \frac{\pi^2}{12} - \frac{9}{8} \ln^2 2 \right), $$

and

$$ \sum_{n=1}^{\infty} (-1)^n \binom{2n}{n} \frac{H_n O_n}{8^n} = \sqrt{\frac{2}{3}} \left( \mathrm{Li}_2(\sqrt{\frac{2}{3}}) - \mathrm{Li}_2(-\sqrt{\frac{2}{3}}) - \frac{\pi^2}{4} - \ln^2 2 + \frac{\ln 2 \ln 3}{2} + \frac{\ln^2 3}{2} \right). $$

*Proof.* To obtain the first series, set $x = -1/4$ in (5.53) and (5.54), and simplify using the identity $G(\frac{1}{4}) = \frac{3}{2} \ln^2 2$. The second series is obtained by setting $x = 1/8$ in (5.53) and (5.54). Note that $G(\frac{1}{8}) = -\ln^2 2 + \frac{i\pi}{2} \ln 2$, and this yields

$$ \sum_{n=1}^{\infty} \binom{2n}{n} \frac{H_n O_n}{8^n} = \sqrt{2} \left( \mathrm{Li}_2(\sqrt{2}) - \mathrm{Li}_2(-\sqrt{2}) - \frac{\pi^2}{4} - \ln^2 2 + \frac{i\pi}{2} \ln 2 \right). $$

To remove the imaginary term, we simplify further using the dilogarithm property (1.4). Finally, insert $x = -1/8$ in (5.53) and simplify making use of the evaluation $G(-\frac{1}{8}) = -\ln^2 2 + \frac{1}{2} \ln 2 \ln 3 + \frac{1}{2} \ln^2 3$.
**Corollary 5.7.** Let $\alpha = (1 + \sqrt{5})/2$ be the golden ratio. Then

$$ \sum_{n=1}^{\infty} \binom{2n}{n} \frac{H_n O_n}{(4\alpha)^n} = \frac{\pi^2}{12}\alpha + \frac{5}{2}\alpha \ln^2 \alpha - 2\alpha \ln 2 \ln \alpha. $$

*Proof.* Set $x = \alpha/4$ in (5.53) and (5.54), and use that $\sqrt{1 - 1/\alpha} = 1/\alpha$ in conjunction with

$$ \mathrm{Li}_2(-\alpha) = -\frac{\pi^2}{10} - \ln^2 \alpha, \quad \mathrm{Li}_2(\alpha) = \frac{7\pi^2}{30} + \frac{1}{2}\ln^2 \alpha - i\pi \ln \alpha, $$

where the second dilogarithm identity follows from (1.4).

**Theorem 5.8.** For all $x \in (-1/4, 1/4)$, we have

$$ \sum_{n=1}^{\infty} \binom{2n}{n} n H_n O_n x^n = \frac{2x}{\sqrt{(1-4x)^3}} \left( \mathrm{Li}_2\left(\frac{1}{\sqrt{1-4x}}\right) - \mathrm{Li}_2\left(-\frac{1}{\sqrt{1-4x}}\right) \right. \\ \left. - \frac{\pi^2}{4} + G(x) + \ln\left(\frac{\sqrt{1-4x}+1}{\sqrt{1-4x}-1}\right) - H(x) \right), \quad (5.56) $$

with $G(x)$ given in (5.54) and $H(x)$ given by

$$ H(x) = \begin{cases} \frac{1}{4x} \left( \ln(1-4x) + 4x \ln\left(-\frac{1-4x}{x}\right) \right), & \text{if } x < 0; \\ \frac{1}{4x} \left( \ln(1-4x) + 4x \left( \ln\left(\frac{1-4x}{x}\right) + i\pi \right) \right), & \text{if } x > 0. \end{cases} \quad (5.57) $$

Alternatively, we can write

$$ \sum_{n=1}^{\infty} \binom{2n}{n} n H_n O_n x^n = \frac{x}{\sqrt{(1-4x)^3}} \left( \mathrm{Li}_2(1-4x) - 4 \mathrm{Li}_2(\sqrt{1-4x}) + \frac{\pi^2}{2} \right. \\ \left. + 2 \ln \left( \frac{\sqrt{1-4x}+1}{\sqrt{1-4x}-1} \right) + \ln \left( \frac{1-4x}{|x|} \right) \left( \ln(1-4x) - 2 \right) - \frac{\ln(1-4x)}{2x} - H(x) \right), $$

where

$$ H(x) = \begin{cases} i\pi \ln(1-4x), & \text{if } x < 0; \\ 2\pi i, & \text{if } x > 0. \end{cases} $$

For $x = 0$ the series on the left-hand side of (5.56) is obviously zero.

*Proof.* Differentiate identity (5.53) with respect to $x$ using the product rule.

Three particular evaluations are offered in the next corollary.

**Corollary 5.9.** We have

$$ \sum_{n=1}^{\infty} (-1)^n \binom{2n}{n} \frac{n H_n O_n}{8^n} = -\frac{\sqrt{6}}{18} \left( \mathrm{Li}_2\left(\sqrt{\frac{2}{3}}\right) - \mathrm{Li}_2\left(-\sqrt{\frac{2}{3}}\right) - \frac{\pi^2}{4} + 2 \ln(\sqrt{2} + \sqrt{3}) \right. \\ \left. - 4 \ln 2 + \ln 3 + \frac{1}{2} \ln 2 \ln 3 - \ln^2 2 + \frac{1}{2} \ln^2 3 \right), $$

$$ \sum_{n=1}^{\infty} \binom{2n}{n} \frac{n H_n O_n}{8^n} = \frac{\sqrt{2}}{2} \left( -\mathrm{Li}_2\left(\frac{1}{\sqrt{2}}\right) - \mathrm{Li}_2(-\sqrt{2}) + \frac{\pi^2}{12} - \frac{9}{8} \ln^2 2 + 2 \ln(1 + \sqrt{2}) \right), $$

$$ \sum_{n=1}^{\infty} \binom{2n}{n} \frac{n H_n O_n}{5^n} = 2\sqrt{5} \left( -\mathrm{Li}_2\left(\frac{1}{\sqrt{5}}\right) - \mathrm{Li}_2(-\sqrt{5}) + \frac{\pi^2}{12} + 2 \ln \alpha + \frac{5}{4} \ln 5 - \frac{1}{8} \ln^2 5 \right), $$

where $\alpha$ is the golden ratio.
We also present a second remarkable evaluation of the complementary series from Corollary 5.7.

**Corollary 5.10.** We have

$$ \sum_{n=1}^{\infty} \binom{2n}{n} \frac{nH_n O_n}{4^n \alpha^n} = \frac{\pi^2 \alpha^2}{24} + \frac{5}{4} \alpha^2 \ln^2 \alpha + \alpha^2 \ln \left( \frac{\alpha^2}{2} \right) + \alpha^3 \ln \alpha - \alpha^2 \ln 2 \ln \alpha. $$

# 6. Series associated with generating function (5.53) and dilogarithm functions

In this section, we derive a second remarkable evaluation of the companion series from Corollary 5.7, along with related series identities rising from the generating function (5.53) and properties of the dilogarithm function.

Writing $(1-x)/4$ for $x$ and substituting $x^2$ for $x$, identity (5.53) can be written for $x \in (0, 1)$ as

$$ x \sum_{n=1}^{\infty} \binom{2n}{n} \frac{H_n O_n}{2^{2n}} (1-x^2)^n = 2 \Re \left( \chi_2 \left( \frac{1}{x} \right) \right) - \frac{\pi^2}{4} + 2 \ln 2 \ln x + 2 \ln^2 x - \ln x \ln(1-x^2), \quad (6.58) $$

where

$$ \chi_2(y) = \frac{1}{2} \mathrm{Li}_2(y) - \frac{1}{2} \mathrm{Li}_2(-y) $$

is Legendre's chi function [Lew81, p. 19] with the known values:

$$ \begin{aligned} \chi_2\left(\tan\frac{\pi}{8}\right) &= \chi_2(\sqrt{2}-1) = \frac{\pi^2}{16} - \frac{1}{4}\ln^2(\sqrt{2}-1), \\ \chi_2(-\beta) &= \frac{\pi^2}{12} - \frac{3}{4}\ln^2\alpha, \qquad \chi_2(-\beta^3) = \frac{\pi^2}{24} - \frac{3}{4}\ln^2\alpha. \end{aligned} $$

**Lemma 6.1.** If $0 \neq u/2 \notin \mathbb{Z}^{-}$, then

$$ \begin{aligned} \int_0^1 x^{u-1} \ln x \, dx &= -\frac{1}{u^2}, & \int_0^1 x^{u-1} \ln^2 x \, dx &= \frac{2}{u^3}, \\ \int_0^1 x^{u-1} \ln x \ln(1-x^2) \, dx &= -\frac{\zeta(2) - H_{u/2}^{(2)}}{2u} + \frac{H_{u/2}}{u^2}, & & (6.59) \end{aligned} $$

$$ \Re \left( \int_0^1 x^{u-1} \chi_2 \left( \frac{1}{x} \right) dx \right) = \frac{2H_u - H_{u/2}}{2u^2} + \frac{\pi^2}{8u}. \quad (6.60) $$

*Proof.* We prove (6.59) and (6.60). A change of variable in (2.18) yields

$$ \int_0^1 x^{u-1} \ln(1-x^2) dx = -\frac{H_{u/2}}{u}, \quad 0 \neq u \in \mathbb{C} \setminus \mathbb{Z}^{-}, $$

which, upon differentiation with respect to $u$, gives (6.59).

A simple change of variable in the integration formula (4.43) gives

$$ \int_0^1 x^{u-1} \mathrm{Li}_2(x^2) dx = \frac{\pi^2}{6u} - \frac{2H_{u/2}}{u^2}. \quad (6.61) $$
Identities (4.43) and (6.61) together with the following functional relations for the dilogarithm:

$$ \mathrm{Li}_2(x) + \mathrm{Li}_2(-x) = \frac{1}{2} \mathrm{Li}_2(x^2), $$

$$ \mathrm{Li}_2\left(\frac{1}{x}\right) + \mathrm{Li}_2(x) = \frac{\pi^2}{3} - \frac{1}{2}\ln^2 x + i\pi \ln x, \quad x < 1, $$

$$ \mathrm{Li}_2\left(-\frac{1}{x}\right) + \mathrm{Li}_2(-x) = -\frac{\pi^2}{6} - \frac{1}{2}\ln^2 x, \quad x > 0, $$

lead directly to the determination of (6.60).

**Theorem 6.2.** We have

$$ \sum_{n=1}^{\infty} \frac{C_n H_n O_n}{2^{2n}} = \frac{\pi^2}{2} + 4 \ln 2, \qquad (6.62) $$

$$ \sum_{n=1}^{\infty} \frac{C_n H_n O_n}{2^{2n}(n+2)} = -\frac{14}{9} + \frac{\pi^2}{6} + \frac{4}{9} \ln 2, \qquad (6.63) $$

$$ \sum_{n=1}^{\infty} \frac{H_n O_n}{(2n+1)(2n+3)} = \frac{1}{2} + \frac{\pi^2}{24} - \frac{1}{2} \ln 2, \qquad (6.64) $$

and more generally, for $0 \neq r \in \mathbb{C} \setminus \mathbb{Z}^{-}$,

$$ \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{(n+(r+1)/2)} \frac{H_n O_n}{2^{2n}} = \frac{r+1}{r^2} \left( 2H_r - 2H_{r/2} - \frac{r}{2} H_{r/2}^{(2)} + \frac{\pi^2}{12}r - 2\ln 2 + \frac{4}{r} \right). \qquad (6.65) $$

*Proof.* Multiply through (6.58) by $x^{r-1}$ and integrate from 0 to 1 applying Lemma 6.1 and the identity

$$ \int_0^1 (1-x^2)^n x^r dx = \frac{\Gamma(n+1)\Gamma((r+1)/2)}{2\Gamma(n+1+(r+1)/2)} = \frac{1}{(r+1)\binom{n+(r+1)/2}{n}}. $$

Identities (6.62), (6.63), and (6.64) follow from evaluating (6.65) at $r = 1$, $r = 3$, and $r = 2$, respectively.

**Corollary 6.3.** If $0 \neq r \in \mathbb{C} \setminus \mathbb{Z}^{-}$, $2r \notin \mathbb{Z}^{-}$, then

$$ \sum_{n=1}^{\infty} \frac{\binom{n+r}{n}}{\binom{2n+2r+1}{2n}} H_n O_n = \frac{2r+1}{4r^2} \left( 2(H_{2r} - H_r) - rH_r^{(2)} + \frac{\pi^2}{6}r - 2\ln 2 + \frac{2}{r} \right). $$

*Proof.* Write $2r$ for $r$ in (6.65) and use $\binom{r+1/2}{s} = \binom{2r+1}{2s} \binom{2s}{s}^{-1} 2^{-2s}$.

**Theorem 6.4.** We have

$$ \sum_{n=1}^{\infty} \frac{C_n H_n H_{n+1} O_n}{2^{2n}} = 16 \ln 2 + \frac{5\pi^2}{3} + 14 \zeta(3), $$

$$ \sum_{n=1}^{\infty} \frac{C_n H_n H_{n+2} O_n}{2^{2n}(n+2)} = -\frac{236}{27} + \frac{28}{27} \ln 2 + \frac{19\pi^2}{54} + \frac{14}{3} \zeta(3), $$

and more generally, for $0 \neq r \in \mathbb{C} \setminus \mathbb{Z}^{-}$,

$$ \begin{aligned} \sum_{n=1}^{\infty} \frac{\binom{2n}{n} H_n H_{n+(r+1)/2} O_n}{2^{2n} \binom{n+(r+1)/2}{n}} = & \frac{4(r+2)}{r^3} (H_r - H_{r/2}) + \frac{4(r+1)}{r^2} H_r^{(2)} - \frac{2r+3}{r^2} H_{r/2}^{(2)} \\ & - \frac{r+1}{r} H_{r/2}^{(3)} - \frac{(2r+1)}{r^2} \frac{\pi^2}{6} + \frac{(r+1)}{r} \zeta(3) - \frac{4(r+2)}{r^3} \ln 2 + \frac{8(2r+3)}{r^4} \qquad (6.66) \\ & + \frac{r+1}{r^2} H_{(r+1)/2} \left( 2H_r - 2H_{r/2} - \frac{r}{2} H_{r/2}^{(2)} + \frac{\pi^2}{12}r - 2\ln 2 + \frac{4}{r} \right). \end{aligned} $$
Proof. Differentiate (6.65) with respect to r to obtain

$$
\begin{aligned}
& \sum_{n=1}^{\infty} \frac{\binom{2n}{n} H_n O_n \left(H_{n+(r+1)/2} - H_{(r+1)/2}\right)}{2^{2n} \binom{n+(r+1)/2}{n}} \\
&= \frac{4(r+2)}{r^3} \left(H_r - H_{r/2}\right) + \frac{4(r+1)}{r^2} H_r^{(2)} - \frac{2r+3}{r^2} H_{r/2}^{(2)} - \frac{r+1}{r} H_{r/2}^{(3)} \\
&\quad - \frac{(2r+1)}{r^2} \frac{\pi^2}{6} + \frac{(r+1)}{r} \zeta(3) - \frac{4(r+2)}{r^3} \ln 2 + \frac{8(2r+3)}{r^4},
\end{aligned}
$$

from which (6.66) follows upon second use of (6.65).

7. A second new generating function

This section deals with the generating function involving the product $C_n H_n O_n$.

**Theorem 7.1.** For all $x \in [-1/4, 1/4)$, we have

$$
\begin{aligned}
\sum_{n=1}^{\infty} C_n H_n O_n x^n ={}& \frac{1}{4x} \left( 2 \operatorname{Li}_2(-\sqrt{1-4x}) - 2 \operatorname{Li}_2(\sqrt{1-4x}) + \frac{\pi^2}{2} \sqrt{1-4x} \right. \\
& \left. + (\ln(1-4x)-4) \ln(1+\sqrt{1-4x}) - \ln(1-4x) \ln(1-\sqrt{1-4x}) \right. \\
& \left. + \sqrt{1-4x} (\ln(1-4x)-2) K(x) + 4 \ln 2 + \frac{\pi^2}{2} \right. \\
& \left. + 2 \sqrt{1-4x} \left( \operatorname{Li}_2\left(-\frac{1}{\sqrt{1-4x}}\right) - \operatorname{Li}_2\left(\frac{1}{\sqrt{1-4x}}\right) - \ln\left(\frac{\sqrt{1-4x}+1}{\sqrt{1-4x}-1}\right) \right) \right)
\end{aligned}
\qquad (7.67)
$$

with $K(x)$ given by

$$
K(x) = \begin{cases} \ln\left(\frac{-x}{1-4x}\right), & \text{if } x < 0; \\ \ln\left(\frac{-x}{1-4x}\right) + i\pi, & \text{if } x > 0. \end{cases} \tag{7.68}
$$

For $x=0$ the series on the left-hand side of (7.67) is obviously zero.

Proof. Combining (1.11) with (1.12) yields

$$
\sum_{n=1}^{\infty} C_n H_n O_n x^n = \frac{1}{2x} \int_{0}^{1} \frac{J(x, t)}{t(1-t)} dt
$$

with

$$
J(x,t) = t - 1 + \sqrt{1-4xt} - \sqrt{1-4x} t - \frac{\sqrt{1-4xt}}{2} \ln(1-4xt) + \frac{\sqrt{1-4x}}{2} \ln(1-4x)t.
$$

We have

$$
\begin{aligned}
I^C(x,t) = \int \frac{J(x,t)}{t(1-t)} dt ={}& \frac{1}{2} \left( 2 \operatorname{Li}_2(-\sqrt{1-4xt}) - 2 \operatorname{Li}_2(\sqrt{1-4xt}) \right. \\
& \left. - 2\sqrt{1-4x} \operatorname{Li}_2\left(-\frac{\sqrt{1-4xt}}{\sqrt{1-4x}}\right) + 2\sqrt{1-4x} \operatorname{Li}_2\left(\frac{\sqrt{1-4xt}}{\sqrt{1-4x}}\right) \right. \\
& \left. - \sqrt{1-4x} (\ln(1-4x) - 2) \ln(1-t) - \ln(1-4xt) \ln(1-\sqrt{1-4xt}) \right. \\
& \left. + (\ln(1-4xt) - 4) \ln(1+\sqrt{1-4xt}) + \sqrt{1-4x} \ln(1-4xt) \ln\left(1-\frac{\sqrt{1-4xt}}{\sqrt{1-4x}}\right) \right. \\
& \left. - \sqrt{1-4x} \ln(1-4xt) \ln\left(1+\frac{\sqrt{1-4xt}}{\sqrt{1-4x}}\right) + 4\sqrt{1-4x} \operatorname{arctanh}\left(\frac{\sqrt{1-4xt}}{\sqrt{1-4x}}\right) \right) + c.
\end{aligned}
$$
From here, we calculate

$$2I^C(x, 1) = 2 \mathrm{Li}_2(-\sqrt{1-4x}) - 2 \mathrm{Li}_2(\sqrt{1-4x}) + \frac{\pi^2}{2}\sqrt{1-4x} - \ln(1-4x)\ln(1-\sqrt{1-4x}) \\ + (\ln(1-4x)-4)\ln(1+\sqrt{1-4x}) - \ln 2\sqrt{1-4x}\ln(1-4x) + 2\ln 2\sqrt{1-4x} \\ + \sqrt{1-4x}\lim_{t \to 1^{-}} \left(- (\ln(1-4x)-2)\ln(1-t) + \ln\left(1-\frac{\sqrt{1-4xt}}{\sqrt{1-4x}}\right)(\ln(1-4xt)-2) \right)$$

with

$$\lim_{t \to 1^{-}} \left( - (\ln(1-4x)-2) \ln(1-t) + \ln\left(1-\frac{\sqrt{1-4xt}}{\sqrt{1-4x}}\right) (\ln(1-4xt)-2) \right) \\ = (\ln(1-4x)-2) \lim_{t \to 1^{-}} \left( \ln\left(\frac{\sqrt{1-4x}-\sqrt{1-4xt}}{1-t}\right) - \frac{1}{2}\ln(1-4x) \right) \\ = (\ln(1-4x)-2) \ln\left(\frac{-2x}{1-4x}\right).$$

By a similar calculation, we get

$$2I^C(x, 0) = -\frac{\pi^2}{2} - 4 \ln 2 + 2\sqrt{1-4x} \ln\left(\frac{\sqrt{1-4x}+1}{\sqrt{1-4x}-1}\right) \\ - 2\sqrt{1-4x} \left( \mathrm{Li}_2\left(-\frac{1}{\sqrt{1-4x}}\right) - \mathrm{Li}_2\left(\frac{1}{\sqrt{1-4x}}\right) \right),$$

where it was used that

$$\lim_{t \to 0} \ln(1 - 4xt) \ln(1 - \sqrt{1 - 4xt}) = 0.$$

Gathering terms and simplifying proves the theorem.

In the following examples, to retain and acknowledge the symmetrical structures, it is even better to keep the purely imaginary terms in the evaluations (which can be canceled out in view of (1.4)).

**Corollary 7.2.** *We have*

$$\sum_{n=1}^{\infty} \frac{(-1)^n}{4^n} C_n H_n O_n = -2 \left( \mathrm{Li}_2(-\sqrt{2}) - \mathrm{Li}_2(\sqrt{2}) \right) - 2\sqrt{2} \left( \mathrm{Li}_2\left(-\frac{1}{\sqrt{2}}\right) - \mathrm{Li}_2\left(\frac{1}{\sqrt{2}}\right) \right) \\ + (4\sqrt{2} - 2\ln 2 + 4) \ln(\sqrt{2} + 1) - \frac{\pi^2}{2}(1 + \sqrt{2}) \\ + (3\sqrt{2}\ln 2 - 6\sqrt{2} - 4) \ln 2 + i\pi \ln 2 \quad (7.69)$$

and

$$\sum_{n=1}^{\infty} (-1)^n \frac{C_n H_n O_n}{8^n} = -4 \left( \mathrm{Li}_2\left(-\sqrt{\frac{3}{2}}\right) - \mathrm{Li}_2\left(\sqrt{\frac{3}{2}}\right) \right) - 2\sqrt{6} \left( \mathrm{Li}_2\left(-\sqrt{\frac{2}{3}}\right) - \mathrm{Li}_2\left(\sqrt{\frac{2}{3}}\right) \right) \\ - \pi^2 \left( \sqrt{\frac{3}{2}} + 1 \right) + 4\sqrt{6} \ln(\sqrt{3} + \sqrt{2}) - 8\ln 2 - 2 \left( \ln\left(\frac{3}{2}\right) - 4 \right) \ln\left(1 + \sqrt{\frac{3}{2}}\right) \quad (7.70) \\ + 2 \ln\left(\frac{3}{2}\right) \left( \ln\left(\sqrt{\frac{3}{2}} - 1\right) + i\pi \right) + \sqrt{6} \left( \ln\left(\frac{3}{2}\right) - 2 \right) (2 \ln 2 + \ln 3).$$
**Remark 7.3.** An alternative version of (7.67) is

$$ \sum_{n=1}^{\infty} C_n H_n O_n x^{n+1} = \frac{\pi^2}{8} + \ln 2 + \frac{1}{4} \left( \begin{aligned} & (1 - \sqrt{1-4x})(\mathrm{Li}_2(1-4x) - 4\mathrm{Li}_2(\sqrt{1-4x})) \\ & - \frac{\pi^2-2}{2}\sqrt{1-4x} - (\ln(1-4x) - 2\sqrt{1-4x})\ln\left(\frac{1-\sqrt{1-4x}}{1+\sqrt{1-4x}}\right) \\ & - 4\ln(1+\sqrt{1-4x}) + \sqrt{1-4x}((\ln(1-4x)-2)\ln(|x|) - (\ln(1-4x)-1)^2) \\ & + \frac{1-\operatorname{sgn}x}{8}i\pi\sqrt{1-4x}(\ln(1-4x)-2) \end{aligned} \right) $$

## 8. Conclusion

In this work, we have explored a wide range of infinite series involving central binomial coefficients, Catalan numbers, and harmonic-type sequences such as $H_n$ and $O_n$. Through analytic techniques, including integral representations and generating functions, we derived several new identities, many of which are not expressible in terms of elementary functions and instead involve special functions like the dilogarithm.

A central contribution of this paper is the evaluation of two nontrivial classes of series

$$ \sum_{n=1}^{\infty} \binom{2n}{n} H_n O_n x^n \quad \text{and} \quad \sum_{n=1}^{\infty} C_n H_n O_n x^n, $$

for which we obtained closed-form expressions or representations involving polylogarithmic terms. These results complement and extend earlier findings by many mathematicians and provide a deeper understanding of the interplay between harmonic numbers and combinatorial coefficients.

While many of the series considered resist expression in closed form, particularly those involving harmonic number products, we have identified special cases–such as those involving the golden ratio– where elegant simplifications are possible. These findings illustrate the rich structure and complexity inherent in such series and open avenues for further investigation, particularly in the context of special values of polylogarithms and their connections to combinatorics.

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Kunle Adegoke

Obafemi Awolowo University, Department of Physics and Engineering Physics
220005 Ile-Ife, Nigeria

*e-mail:* adegoke00@gmail.com

Robert Frontczak

Independent Researcher

72764 Reutlingen, Germany

*e-mail:* robert.frontczak@web.de

Taras Goy

Vasyl Stefanyk Precarpathian National University, Faculty of Mathematics and Computer Science
76018 Ivano-Frankivsk, Ukraine

*e-mail:* taras.goy@pnu.edu.ua