# Infinite products involving the period-doubling sequence

John Campbell

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# Infinite products involving the period-doubling sequence

JOHN M. CAMPBELL

## Abstract

We explore the evaluation of infinite products involving the automatic sequence $(d_n : n \in \mathbb{N}_0)$ known as the period-doubling sequence, inspired by the work of Allouche, Riasat, and Shallit on the evaluation of infinite products involving the Thue-Morse or Rudin-Shapiro sequences. Our methods allow for the application of integral operators that result in new product expansions for expressions involving the dilogarithm function, resulting in new formulas involving Catalan's constant $G$, such as the formula

$$ \prod_{n=1}^{\infty} \left( \left( \frac{n+2}{n} \right)^{n+1} \left( \frac{4n+3}{4n+5} \right)^{4n+4} \right)^{\frac{d_n}{n}} = \frac{e^{\frac{2G}{\pi}}}{\sqrt{2}} $$

introduced in this article. More generally, the evaluation of infinite products of the form $\prod_{n=1}^{\infty} e(n)^{d_n}$ for an elementary function $e(n)$ is the main purpose of our article. Past work on infinite products involving automatic sequences has mainly concerned products of the form $\prod_{n=1}^{\infty} R(n)^{a(n)}$ for an automatic sequence $a(n)$ and a rational function $R(n)$, in contrast to our results as in above displayed product evaluation. Our method also allows us to obtain new evaluations involving $\frac{\zeta(3)}{\pi^2}$ for infinite products involving the period-doubling sequence.

## 1 Introduction

We may let the period-doubling sequence be denoted as $(d_n : n \in \mathbb{N}_0)$ and defined in the following manner. For $n \in \mathbb{N}_0$, we may set $d_n$ as the highest power of 2 dividing $n+1$, modulo 2, giving us an automatic sequence given by the fixed point of the morphism that is on the free monoid on the alphabet $\{0, 1\}$ and that satisfies $0 \mapsto 01$ and $1 \mapsto 00$. For example, the highest power of 2 dividing $9+1 = 2 \times 5$ is given by the exponent of $2^1$, so that $d_9 = 1 \pmod{2}$, and this agrees with (1) below. The integer sequence

$$ (d_n : n \in \mathbb{N}_0) = (0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, \dots) \quad (1) $$

may be regarded as an important instance of an automatic sequence, and this motivates the exploration of new properties concerning the integer sequence in (1). Motivated by the infinite products that are given in [9] and that involve the Thue-Morse and Rudin-Shapiro sequences, we pursue an exploration of the evaluation of infinite products involving the period-doubling sequence.

The period-doubling sequence may be defined recursively so that $d(2n) = 0$ and $d(4n + 1) = 1$ and $d(4n + 3) = d_n$ for $n \in \mathbb{N}_0$. So, for a sequence $(R(n) : n \in \mathbb{N}_0)$ that we typically set as a rational function, we may apply the recursive definition for $d$ to show that

$$ \prod_{n=1}^{\infty} \left( \frac{R(n)}{R(4n+3)} \right)^{d_n} = \prod_{n=0}^{\infty} R(4n+1), \quad (2) $$
under the assumption that both of the infinite products shown above are convergent. With regard to the identity in (2), a similar approach was applied in [9] to evaluate series of the forms

$$ \prod_{n=1}^{\infty} S(n)^{(-1)^{t_n}} \quad \text{and} \quad \prod_{n=1}^{\infty} S(n)^{(-1)^{v_n}}, \qquad (3) $$

for various rational functions $S(n)$, and where $(t_n : n \in \mathbb{N}_0)$ and $(v_n : n \in \mathbb{N}_0)$ respectively denote the Thue–Morse and Rudin–Shapiro sequences (defined below). The research due to Allouche et al. on the symbolic evaluation of infinite products as in (3) motivates the determination of analogous results for automatic sequences other than $(t_n : n \in \mathbb{N}_0)$ and $(v_n : n \in \mathbb{N}_0)$.

In view of the identity in (2), we set the sequence $R(n)$ as a rational function involving free parameters such that both sides of (2) converge for special values of the input parameters. It was known to Euler that convergent infinite products of rational functions admit evaluations in terms of the $\Gamma$-function [13] that is defined for $\Re(x) > 0$ via the Euler integral $\Gamma(x) = \int_0^\infty u^{x-1}e^{-u} du$ [22, §8]. However, to employ the integration approach described below, we would want to determine values of the aforementioned parameters so that right-hand product in (2) reduces to an elementary function, through the use of the famous reflection formula (see [22, §17 & §20])

$$ \Gamma(x)\Gamma(1-x) = \frac{\pi}{\sin(\pi x)}, \qquad (4) $$

or via Euler's product formula for the sine function (see [14, 17], for example):

$$ \frac{\sin x}{x} = \prod_{j=1}^{\infty} \left( 1 - \frac{x^2}{j^2 \pi^2} \right). \qquad (5) $$

So, if we determine a rational function $R(n)$ such that (2) reduces to an equality of the form

$$ \prod_{n=1}^{\infty} \left( \frac{R(n)}{R(4n+3)} \right)^{d_n} = e(n) \qquad (6) $$

for an elementary function $e(n)$, we would proceed to rewrite (6) as

$$ \sum_{n=1}^{\infty} d_n \ln \left( \frac{R(n)}{R(4n+3)} \right) = \ln(e(n)), \qquad (7) $$

and then apply integral operators to both sides of (7) (possibly after a change of variables or some other manipulation of the equality in (7)), and then argue, if possible, that one may reverse the order of integration and infinite summation. Ideally, the summand of the resultant series may be written in a natural way as a logarithmic expression, and the integral involving the right-hand side of (7) would admit an explicit symbolic form, so that we may obtain a new infinite product evaluation. Our interest in the integration of (7) is partly due to how logarithmic-trigonometric integrals often yield evaluations in terms of the
polylogarithm $Li_n(z) = 1 + \frac{z^2}{2n} + \frac{z^3}{3n} + \cdots$ or fundamental constants such as Catalan's constant $G = 1 - \frac{1}{3^2} + \frac{1}{5^2} - \cdots$.

It is only in exceptional cases and for very special choices of the rational function $R(n)$ that we may successfully apply the above described evaluation technique, which is representative of the remarkable nature of the result highlighted in Section 1.1 below. The main results in this article were discovered in an experimental fashion, with the use of searches via Computer Algebra System (CAS) software for rational functions $R(n)$ that allow us to go through with the above indicated procedure.

Research as in [23, 25] involving infinite series containing the period-doubling sequence has inspired much about our applications in Section 2 of infinite sums involving the sequence in (1). See also references such as [15, 19] for research based on the period-doubling sequence, along with the many references given in the entry indexed as A096268 in the On-line Encyclopedia of Integer Sequences. There is a rich history associated with the study of the closed-form evaluation of infinite products with automatic sequences involved as exponents. In this regard, references as in [5, 6, 9, 11, 20] are relevant to our work and have inspired much about the new results introduced in Section 2 below.

## 1.1 A motivating result

We highlight as a main motivating example the new formula

$$ \prod_{n=1}^{\infty} \left( \left( \frac{n+2}{n} \right)^{n+1} \left( \frac{4n+3}{4n+5} \right)^{4n+4} \right)^{d_n} = \frac{e^{\frac{2G}{\pi}}}{\sqrt{2}} \quad (8) $$

proved in this article through the evaluation technique described above. There are many reasons as to the research interest in the equality in (8), as we describe below.

Let the Thue-Morse sequence be denoted as $(t_n : n \in \mathbb{N}_0)$ and defined so that $t_n$ equals the sum, modulo 2, of the binary digits of $n$. Previous results on infinite products involving automatic sequences, such as the Woods-Robbins product formula [24, 26] (cf. [7, 8, 9])

$$ \prod_{n=0}^{\infty} \left( \frac{2n+1}{2n+2} \right)^{(-1)^{t_n}} = \frac{1}{\sqrt{2}}, \quad (9) $$

have mainly been given by identities for infinite products of the form

$$ \prod_{n} R(n)^{a(n)} \quad (10) $$

for a rational function $R(n)$ and an automatic sequence $a(n)$. In contrast, the formula in (8) is for an infinite product of the form

$$ \prod_{n} f(n)^{a(n)} \quad (11) $$
for an *elementary function* $f(n)$ that assumes rational values for $n \in \mathbb{N}$, as opposed to a rational function. It seems that there is not much known about the evaluation of products of the form (11) if we let $f(n)$ be as before, if we consider the main article that has inspired our results [9] and the references therein, including [5, 7, 8]. Closed-form evaluations for expressions as in (10) that are included in Theorems in [9] include

$$ \prod_{n=0}^{\infty} \left( \frac{(4n+1)(4n+4)}{(4n+2)(4n+3)} \right)^{t_n} = \frac{\pi^{3/4} \sqrt{2}}{\Gamma\left(\frac{1}{4}\right)} $$

and

$$ \prod_{n=0}^{\infty} \left( \frac{(8n+1)(8n+7)}{(8n+3)(8n+5)} \right)^{t_n} = \sqrt{2\sqrt{2}-2} \qquad (12) $$

and

$$ \prod_{n=0}^{\infty} \left( \frac{4(n+2)(2n+1)^3(2n+3)^3}{(n+3)(n+1)^2(4n+3)^4} \right)^{(-1)^{v_n}} = 1, \qquad (13) $$

where $(v_n : n \in \mathbb{N}_0)$ denotes the Rudin–Shapiro sequence such that $v_{2n} = v_n$ and $v_{4n+1} = v_n$ and $v_{4n+3} = 1 - v_{2n+1}$ with $v_0 = 0$. The above evaluations from [9] further motivate the research interest in our product evaluations as in (8).

The new formula in (8) is also of interest in terms of it relates, via product of rational numbers, the fundamental constants $e$, $G$, $\pi$, and $\sqrt{2}$, in a manner not unlike the famous formula $e^{\pi i} + 1 = 0$. Also, the research interest pertaining to the new result displayed in (8) is partly given by how this infinite product evaluation relates to product formulas as in [3], such as the evaluation

$$ \frac{4}{\pi} = \prod_{n=2}^{\infty} \left(1 + \frac{1}{n+1}\right)^{2\rho(n)[\ln_2(n)-1]}, $$

where the 4-periodic sequence $\rho$ is such that $\rho(0) = 1$, $\rho(1) = -1$, and $\rho(2) = \rho(3) = 0$. We are to prove results of that are of similar nature relative to (8) and that involve a linear combination of $\frac{2G}{\pi}$ and $\frac{\zeta(3)}{\pi^2}$, which recall the product formulas in [2, 4] related to product formulas as in

$$ \lim_{n \to \infty} \prod_{k=1}^{2n+1} e^{-1/4} \left(1 - \frac{1}{k+1}\right)^{\frac{k(k+1)}{2}(-1)^k} = \exp\left(\frac{7\zeta(3)}{4\pi^2} + \frac{1}{4}\right) $$

and

$$ \lim_{n \to \infty} \prod_{k=1}^{2n} \left(1 - \frac{2}{2k+1}\right)^{k(-1)^k} = \exp\left(\frac{2G}{\pi} - \frac{1}{2}\right) $$

given by Kachi and Tzermias in [18].
## 2 Main results

Setting

$$R(n) = 1 - \frac{1}{an^2 + bn + c} \qquad (14)$$

in (2), the right-hand side of (2) reduces to

$$\frac{a+b+c-1}{a+b+c} \Gamma \left[ \frac{-\sqrt{b^2-4ac+10a+b}}{8a}, \frac{\sqrt{b^2-4ac+10a+b}}{8a} ; \frac{-\sqrt{-4ac+4a+b^2+10a+b}}{8a}, \frac{\sqrt{-4ac+4a+b^2+10a+b}}{8a} \right], \qquad (15)$$

writing

$$\Gamma \left[ \begin{matrix} \alpha, \beta, \dots, \gamma \\ A, B, \dots, C \end{matrix} \right] = \frac{\Gamma(\alpha)\Gamma(\beta)\dots\Gamma(\gamma)}{\Gamma(A)\Gamma(B)\dots\Gamma(C)}. $$

The experimental use of the Mathematica CAS has led us to discovery a variety of combinations of values for the input parameters $a$, $b$, and $c$ such that the desired conditions for (2) described in Section 1 hold, as below. With regard to the upper $\Gamma$-arguments in (15), by setting these arguments to both be equal to $\frac{3}{2}$ in the hope of obtaining an evaluation involving $\Gamma^2(\frac{3}{2}) = \frac{\pi}{4}$ and satisfying the desired conditions described in Section 1, we proceed to solve for $a$, $b$, and $c$, via the Mathematica input reproduced below, noting that the upper arguments in (15) cannot be set to lower half-integer values due to convergence issues with the infinite product on the left-hand side of (2).

$$\text{Solve}[\{(10 a + b - \text{Sqrt}[b^2 - 4 a c])/(8 a) == 3/2, (10 a + b + \text{Sqrt}[b^2 - 4 a c])/(8 a) == 3/2\}, \{a, b, c\}]$$

This gives us that $b = 2a$ and $c = a$, and this, in turn, has led us to formulate our proof of the following result.

**Lemma 1.** For a free parameter $a$, the equality

$$\cos \left( \frac{\pi}{4\sqrt{a}} \right) = \prod_{n=1}^{\infty} \left( \frac{16(a + 2an + an^2 - 1)}{16a + 32an + 16an^2 - 1} \right)^{d_n} \qquad (16)$$

holds if the above infinite product converges.

*Proof.* As suggested above, we are to set $b = 2a$ and $c = a$ in the rational function in (14). So, the identity in (2) then gives us that the infinite product in (16) equals the expression in (15), again subject to $b = 2a$ and $c = a$. Setting $b = 2a$ and $c = a$ in (15), we obtain

$$\frac{4a-1}{4a} \Gamma \left[ \frac{\frac{3}{2}}{\frac{12a-2\sqrt{a}}{8a}}, \frac{\frac{3}{2}}{\frac{12a+2\sqrt{a}}{8a}} \right].$$

We may rewrite the above expression as

$$\frac{(4a - 1)\pi}{16a\Gamma\left(3 - \frac{2\sqrt{a}+12a}{8a}\right)\Gamma\left(\frac{2\sqrt{a}+12a}{8a}\right)},$$

so that repeated applications of the reflection formula in (4) give us the desired result. $\square$
Although we are to apply Lemma 1 to construct a full proof of a result related to the main result highlighted in Section 1.1, special cases of Lemma 1 are of interest in their own right and nicely relate to the work of Allouche, Riasat, and Shallit in [9].

**Example 1.** Setting $a = 4$ gives us the closed form

$$ \prod_{n=1}^{\infty} \left( \frac{16(2n+1)(2n+3)}{(8n+7)(8n+9)} \right)^{d_n} = \frac{\sqrt{2+\sqrt{2}}}{2}, $$

which is very reminiscent of the Allouche–Riasat–Shallit evaluation highlighted as part of a Theorem in [9] and reproduced in (12).

**Example 2.** Setting $a = \frac{9}{4}$ gives us the closed form

$$ \prod_{n=1}^{\infty} \left( \frac{4(3n+1)(3n+5)}{(6n+5)(6n+7)} \right)^{d_n} = \frac{\sqrt{3}}{2}. $$

Lemma 1 is the main key to our below proof.

**Theorem 1.** *The product evaluation*

$$ \prod_{n=1}^{\infty} \left( 16 \frac{(n+2)^{n+2}}{n^n} \frac{(4n+3)^{4n+3}}{(4n+5)^{4n+5}} \right)^{d_n} = \frac{1}{2} e^{\frac{2G}{\pi}} $$

holds true.

*Proof.* We enforce the substitution $a = \frac{1}{\alpha^2}$ in Lemma 1, so that we may obtain that

$$ \cos\left(\frac{\pi\alpha}{4}\right) = \prod_{n=1}^{\infty} \left( \frac{16(1+n-\alpha)(1+n+\alpha)}{(4+4n-\alpha)(4+4n+\alpha)} \right)^{d_n} $$

for positive $\alpha$, which gives us that

$$ \ln\left(\cos\left(\frac{\pi\alpha}{4}\right)\right) = \sum_{n=1}^{\infty} d_n \ln\left(\frac{16(1+n-\alpha)(1+n+\alpha)}{(4+4n-\alpha)(4+4n+\alpha)}\right). \quad (17) $$

An antiderivative with respect to $\alpha$ of the left-hand side is

$$ \frac{1}{8}i\pi\alpha^2 - \alpha \ln\left(1+e^{\frac{i\pi\alpha}{2}}\right) + \alpha \ln\left(\cos\left(\frac{\pi\alpha}{4}\right)\right) + \frac{2i\mathrm{Li}_2\left(-e^{\frac{i\pi\alpha}{2}}\right)}{\pi}. $$

An antiderivative with respect to $\alpha$ of the summand of the series in (17) is as below:

$$ d_n \left( \alpha \ln \left( \frac{16(-\alpha + n + 1)(\alpha + n + 1)}{(-\alpha + 4n + 4)(\alpha + 4n + 4)} \right) + $$
$$2(n+1) \left( \tanh^{-1} \left( \frac{\alpha}{n+1} \right) - 4 \tanh^{-1} \left( \frac{\alpha}{4n+4} \right) \right).$$

By the Dominated Convergence Theorem, we may interchange the operators $\int_0^1 \cdot d\alpha$ and $\sum_{n=0}^{\infty}$, so that the evaluation of the limits as $\alpha \to 0$ and as $\alpha \to 1$ of the above antiderivatives gives us an equivalent formulation of the desired result. □

By setting the expression $R(4n+1)$ in (2) so as to agree with previously known products that admit closed forms, this gives us a direct way of generating closed-form identities for products involving $d_n$, but we instead consider the more challenging situation given by exchanging limiting operations as in our proof of Theorem 1. For example, as in [11], we record the classical result

$$\prod_{n=1}^{\infty} \frac{(2n)(2n)}{(2n-1)(2n+1)} = \frac{\pi}{2} \qquad (18)$$

known as Wallis' product formula, and we find that the identity in (2) allows us in a direct way to transform (18) by writing

$$\prod_{n=1}^{\infty} \left( \frac{4(n+2)(n+3)^2}{(n+5)(2n+3)^2} \right)^{d_n} = \frac{\pi}{2}, \qquad (19)$$

but the immediate way in which (19) was obtained is in contrast to what was required in our proof of Theorem 1, which leads us to consider variants and generalizations of (8).

**Lemma 2.** *The equality*

$$\sqrt{2} \cos \left( \frac{\pi}{4 \sqrt{\frac{b}{b+2}}} \right) = \prod_{n=1}^{\infty} \left( \frac{(4n+3)(4n+5)(bn^2 + 2bn - 2)}{n(n+2)(16bn^2 + 32bn + 15b - 2)} \right)^{d_n}$$

holds for positive b.

*Proof.* By setting $a = \frac{b}{2}$ and $c = 0$ in (2) and (15), we may employ the same argument as in the proof of Lemma 1. □

Although we are mainly interested in Lemma 2 because of our applying this Lemma to prove a variant of (8) via the argument employed in the proof of Theorem 1, special cases of Lemma 2 are of interest in their own right, especially in view of evaluations from [9].

**Example 3.** Setting $b = \frac{18}{7}$ in Lemma 2, we may obtain that

$$\prod_{n=1}^{\infty} \left( \frac{(3n-1)(3n+7)(4n+3)(4n+5)}{16n(n+2)(3n+2)(3n+4)} \right)^{d_n} = \frac{1}{\sqrt{2}},$$

which strongly recalls the same evaluation for the Woods-Robbins product formula in (9).
**Example 4.** Setting $b = -\frac{8}{3}$ in Lemma 2, we may obtain that

$$ \prod_{n=1}^{\infty} \left( \frac{(2n+1)(2n+3)(4n+3)(4n+5)}{n(n+2)(8n+7)(8n+9)} \right)^{d_n} = \sqrt{\frac{2+\sqrt{2}}{2}}, $$

which recalls the formula in (12) due to Allouche, Riasat, and Shallit [9].

**Theorem 2.** The formula in (8) holds true.

*Proof.* We set $b$ as $\frac{2}{\beta^2-1}$, giving us, from Lemma 2, that

$$ \sqrt{2} \cos \left( \frac{\pi \beta}{4} \right) = \prod_{n=1}^{\infty} \left( \frac{(4n+3)(4n+5)(n^2+2n+1-\beta^2)}{n(n+2)(16n^2+32n+16-\beta^2)} \right)^{d_n} \quad (20) $$

for positive $\beta$. Taking the natural logarithm of both sides of this equality, we obtain a series expansion for $\ln(\sqrt{2} \cos(\frac{\pi\beta}{4}))$. An antiderivative for this expression, with respect to $\beta$, is as below:

$$ \frac{1}{8}i\pi\beta^2 - \beta \ln\left(1 + e^{\frac{i\pi\beta}{2}}\right) + \beta \ln\left(\sqrt{2}\cos\left(\frac{\pi\beta}{4}\right)\right) + \frac{2i\mathrm{Li}_2\left(-e^{\frac{i\pi\beta}{2}}\right)}{\pi} $$

An antiderivative of

$$ \ln \left( \frac{(4n + 3)(4n + 5)(n^2 + 2n + 1 - \beta^2)}{n(n + 2)(16n^2 + 32n + 16 - \beta^2)} \right) $$

is

$$ \beta \ln \left( \frac{(4n + 3)(4n + 5)(n^2 + 2n + 1 - \beta^2)}{n(n + 2)(16n^2 + 32n + 16 - \beta^2)} \right) + 2(n + 1) \tanh^{-1} \left( \frac{\beta}{n + 1} \right) - 8(n + 1) \tanh^{-1} \left( \frac{\beta}{4n + 4} \right), $$

and we may thus follow the same line of reasoning as in the proof of Theorem 1. $\square$

Our proof of Theorem 2 has led us to experimentally discovery the following result involving Apéry's constant $\zeta(3) = 1 + \frac{1}{2^3} + \frac{1}{3^3} + \cdots$.

**Theorem 3.** The evaluation

$$ \prod_{n=1}^{\infty} \left( \frac{(4n + 3)^{16}(4n + 5)^{16}}{2^{64}n(n + 1)^{30}(n + 2)} \right)^{\frac{d_n(n+1)^2}{2}} = \frac{e^{\frac{2G}{\pi} - \frac{21\zeta(3)}{8\pi^2}}}{\sqrt[4]{2}} $$

holds true.
Proof. We again apply the natural logarithm to (21), and we proceed to multiply both sides by $\beta$:

$$ \beta \ln \left( \sqrt{2} \cos \left( \frac{\pi \beta}{4} \right) \right) = \sum_{n=1}^{\infty} \beta d_n \ln \left( \frac{(4n+3)(4n+5)(n^2+2n+1-\beta^2)}{n(n+2)(16n^2+32n+16-\beta^2)} \right). \quad (21) $$

An antiderivative for the left-hand side is as below:

$$ \frac{1}{24} i \pi \beta^3 - \frac{1}{2} \beta^2 \ln \left(1 + e^{\frac{i \pi \beta}{2}}\right) + \frac{1}{2} \beta^2 \ln \left(\sqrt{2} \cos \left(\frac{\pi \beta}{4}\right)\right) + \frac{2i\beta \mathrm{Li}_2\left(-e^{\frac{i\pi\beta}{2}}\right)}{\pi} - \frac{4\mathrm{Li}_3\left(-e^{\frac{i\pi\beta}{2}}\right)}{\pi^2} $$

An antiderivative of the summand of the above series is

$$ \frac{1}{2} d_n \left( \beta^2 \ln \left( \frac{(4n+3)(4n+5)(n^2+2n+1-\beta^2)}{n(n+2)(16n^2+32n+16-\beta^2)} \right) - (n+1)^2 \left( \ln \left( (n+1)^2 - \beta^2 \right) - 16 \ln \left( 16(n+1)^2 - \beta^2 \right) \right) \right). $$

Again, we may reverse the order of integration and infinite summation using the Dominated Convergence Theorem. □

**Theorem 4.** The evaluation

$$ \prod_{n=1}^{\infty} \left( \frac{\left( \frac{(4n+3)(4n+5)}{16} \right)^{(4n+3)(4n+5)}}{(n+1)^{30(n+1)^2} (n(n+2))^{n(n+2)}} \right)^{d_n} = \frac{1}{2} e^{\frac{4G}{\pi} - \frac{21\zeta(3)}{4\pi^2}} $$

holds true.

*Proof.* From (17), we may obtain that

$$ \alpha \ln \left( \cos \left( \frac{\pi \alpha}{4} \right) \right) = \sum_{n=1}^{\infty} d_n \alpha \ln \left( \frac{16(1+n-\alpha)(1+n+\alpha)}{(4+4n-\alpha)(4+4n+\alpha)} \right). $$

An antiderivative for the left-hand side is

$$ \frac{1}{24} i \pi \alpha^3 - \frac{1}{2} \alpha^2 \ln \left(1 + e^{\frac{i \pi \alpha}{2}}\right) + \frac{1}{2} \alpha^2 \ln \left(\cos \left(\frac{\pi \alpha}{4}\right)\right) + \frac{2i\alpha \mathrm{Li}_2\left(-e^{\frac{i\pi\alpha}{2}}\right)}{\pi} - \frac{4\mathrm{Li}_3\left(-e^{\frac{i\pi\alpha}{2}}\right)}{\pi^2}. $$

An antiderivative for the summand of the above series is as below:

$$ \frac{1}{2} d_n \left( \alpha^2 \ln \left( \frac{16(-\alpha + n + 1)(\alpha + n + 1)}{(-\alpha + 4n + 4)(\alpha + 4n + 4)} \right) - (n+1)^2 \left( \ln \left( (n+1)^2 - \alpha^2 \right) - 16 \ln \left( 16(n+1)^2 - \alpha^2 \right) \right) \right). $$

Taking the appropriate limits, with $\alpha \to 0$ and $\alpha \to 1$, we may again argue that it is possible to interchange the order of the application of the operators $\int_0^1 \cdot d\alpha$ and $\sum_{n=0}^{\infty} \cdot$. □
## 2.1 Log-sine integrals

By setting $R(n) = 1 - \frac{16x^2}{(n+3)^2}$ in (2), an equivalent formulation of the Euler product formula in (5) gives us, via (2), that

$$ \prod_{n=1}^{\infty} \left( \frac{(2n+3)^2(n-4x+3)(n+4x+3)}{(n+3)^2(2n-2x+3)(2n+2x+3)} \right)^{d_n} = \frac{\sin(\pi x)}{\pi x} \quad (22) $$

for suitably bounded $x$. There is a great amount of literature on the study of log-sine integrals in the form of Clausen's integral $Cl_2(\theta) = -\int_0^\theta \ln(2\sin(\frac{t}{2})) dt$ and the special function given by *log-sine integrals* of the form

$$ Ls_n^{(k)}(\sigma) := - \int_0^\sigma \theta^k \ln^{n-1-k} \left| 2 \sin \frac{\theta}{2} \right| d\theta, \quad (23) $$

as defined in [12]. The work of Borwein and Straub on integrals of the form indicated in (23) [12] leads us to apply the series identity

$$ \sum_{n=1}^{\infty} d_n \ln \left( \frac{(2n+3)^2(n-4x+3)(n+4x+3)}{(n+3)^2(2n-2x+3)(2n+2x+3)} \right) = \ln \left( \frac{\sin(\pi x)}{\pi x} \right) \quad (24) $$

that we derive from (22), following our approach given by our proofs of Theorems 1, 2, 3, and 4. In particular, the following result is of interest because of how closely the infinite product evaluation resembles that for the notable product formula

$$ \prod_{n=1}^{\infty} \left( 1 + \frac{2}{n} \right)^{(-1)^{n+1}n} = \frac{\pi}{2e} \quad (25) $$

highlighted in the Wolfram Mathworld entry on infinite products and included, via its decimal expansion, in the OEIS (see entry A086056 in the OEIS and the references therein [1, 21]).

**Theorem 5.** *The closed-form evaluation*

$$ \prod_{n=1}^{\infty} \left( \frac{16(n+3)^4(n+2)^{4n+8}}{(n+1)^{3n+3}(n+5)^{n+5}(2n+3)^4} \right)^{d_n} = \frac{\pi^2}{e^2} $$

holds true.

*Proof.* By rewriting (24) so that

$$ \sum_{n=1}^{\infty} d_n \ln \left( \frac{(3+2n)^2(3+n-2u)(3+n+2u)}{(3+n)^2(3+2n-u)(3+2n+u)} \right) = \ln \left( \frac{2 \sin \left( \frac{\pi u}{2} \right)}{\pi u} \right), $$

we find that an antiderivative for the right-hand side is as below:

$$ \frac{i \left( \frac{\pi^2 u^2}{4} + \mathrm{Li}_2 \left( e^{i \pi u} \right) \right)}{\pi} + u + u \left( -\ln \left( 1 - e^{i \pi u} \right) \right) + u \ln \left( \frac{2 \sin \left( \frac{\pi u}{2} \right)}{\pi u} \right). $$
An antiderivative for the above summand is as below:

$$d_n \left( u \ln \left( \frac{(2n + 3)^2 (n^2 + 6n - 4u^2 + 9)}{(n + 3)^2 (4n^2 + 12n - u^2 + 9)} \right) + (n + 3) \tanh^{-1} \left( \frac{2u}{n + 3} \right) - 2(2n + 3) \tanh^{-1} \left( \frac{u}{2n + 3} \right) \right).$$

The desired result then follows by taking the required limits and reversing the order of the operators $\int_0^1 du$ and $\sum_{n=1}^\infty \cdot$ $\square$

We may obtain many further results of a similar nature, relative to Theorem 5, using variants of (24) and of our proof of Theorem 5. We encourage the exploration of this area, especially in view of the closely related evaluation given in (25) and in [1, 21].

## 3 Conclusion

By employing series rearrangement identities such as the modified Abel lemma on summation by parts, but with the use of logarithmic expressions as input, this gives us a way of manipulating identities as in (2) so as to obtain new results on automatic sequences related to $(d_n : n \in \mathbb{N}_0)$. We leave it to as separate project to explore this idea in conjunction with the methods applied in Section 2.

Our work has been focused on infinite products involving the period-doubling sequence, but our methods seem to be broadly applicable to automatic sequences other than $(d_n : n \in \mathbb{N}_0)$. How can we determine infinite product evaluations as in Theorems 1–3 using automatic sequences other than $(d_n : n \in \mathbb{N}_0)$?

Ideally, we would want to systematically generalize our methods so as to be applicable to general classes of automatic sequences. In this regard, the results of Hu in [16] may be relevant, in consideration as to how Hu in [16] generalized results in [10] to sequences $a_{w,B}(n)$ counting the occurrences of a fixed word $w$ in the base-$B$ expansion of $n \in \mathbb{N}$, so as to obtain product formulas as in

$$\prod_{n=1}^{\infty} \left( \frac{9n + 7}{9n + 8} \right)^{(-1)^{a_{21,3}(n)}} = \frac{8}{7\sqrt{3}}.$$

Recall the identity in (2) that served as a starting point for our main results. These results were obtained by determining input functions $R$ such that the right-hand product in (2) admitted an explicit evaluation as an elementary function. One may want to instead consider a “backwards” approach, by trying to insist that the expression $\frac{R(n)}{R(4n+3)}$ in the given product of expressions of the form $\left(\frac{R(n)}{R(4n+3)}\right)^{d_n}$ is equal to a given function and then trying to solve for $R$, instead of starting with $R$ as before. For example, in the hope of obtaining a direct analogue of the Woods–Robbins product in (9), this leads us to set

$$\frac{R(n)}{R(4n+3)} = \frac{2(n-1)+1}{2(n-1)+2} \qquad (26)$$
but CAS software does not seem to be able to solve for $R$ in (26). For example, inputting

$$RSolve[R[n]/R[4 n + 3] == ((2 (n - 1) + 1)/(2 (n - 1) + 2)), R[n], n]$$

into Mathematica, this CAS is not able to provide any evaluation for $R$. Experimenting with how $R$ may be expressed in terms of special functions, for a function $R$ satisfying relations as in (26), may be worthwhile.

## Acknowledgements

The author is very thankful to Dr. Jean-Paul Allouche for many useful comments related to this article.

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John M. Campbell

Toronto Metropolitan University

350 Victoria St, Toronto, ON M5B 2K3

jmaxwellcampbell@gmail.com