Principal Solutions of Recurrence Relations and Irrationality Questions in Number Theory ^†

Abstract

We apply the theory of disconjugate linear recurrence relations to the study of irrational quantities in number theory. In particular, for an irrational number associated with solutions of linear three-term recurrence relations, we show that there exists a four-term linear recurrence relation whose solutions show that the number has an irrational square if and only if the four-term recurrence relation has a principal solution of a certain type. The result is extended to higher-order recurrence relations, and a transcendence criterion can also be formulated in terms of these principal solutions. The method generates new series expansions of positive integer powers of $\zeta \left(3\right)$ and $\zeta \left(2\right)$ in terms of Apéry’s now classic sequences.

1. Introduction

Of the methods used today to test for the irrationality of a given number, we cite two separate approaches, one which seems to have overtaken the other recently. The first method is a direct consequence of Apéry’s landmark paper [1], which uses two independent solutions of a specific three-term recurrence relation (see (18) below) to generate a series of rationals whose limit at infinity is $\zeta \left(3\right)$. Many new proofs and surveys of such arguments have appeared since, e.g., Beukers [2], Nesterenko [3], Fischler [4], Cohen [5], Murty [6], Badea [7], and Zudilin [8,9,10] to mention a few in a list that is far from exhaustive.

The idea and the methods used in Apéry’s work [1] were since developed and have produced results such as André-Jeannin’s proof of the irrationality of the inverse sum of the Fibonacci numbers [11], along with a special inverse sum of Lucas numbers [12], and Zudilin’s derivation [13] of a three-term recurrence relation for which there exists two rational valued solutions whose quotients approach Catalan’s constant. In addition, we cite Zudilin’s communication [14] of a four-term recurrence relation (third-order difference equation) for which there exist solutions whose quotients converge to $\zeta \left(5\right)$, but no irrationality results are derived.

One approach involves consideration of the vector space V over $\mathbb{Q}$ spanned by 1, $\zeta \left(3\right)$, $\zeta \left(5\right)$, …, $\zeta (2n+1)$. Using a criterion by Nesterenko [15] on the linear independence of a finite number of reals, Rivoal [16] proved that $dimV\ge clogn$ for all sufficiently large n, from which it follows that the list $\zeta \left(3\right),\zeta \left(5\right),\dots$ contains infinitely many irrationals. Rivoal [17] complements this result by showing that at least one of the numbers $\zeta \left(5\right),\zeta \left(7\right),\dots ,\zeta \left(21\right)$ is irrational. In the same vein, Zudilin [18] shows that at least one of $\zeta \left(5\right),\zeta \left(7\right),\zeta \left(9\right),\zeta \left(11\right)$ is irrational. Recently, Rival and Zudilin [19] showed that among the odd zeta values $\zeta \left(3\right),\zeta \left(5\right),\zeta \left(7\right),\dots ,\zeta \left(63\right)$, at least two of them must be irrational. The latter result was complemented by Lai and Yu’s theorem [20] that gives an effective lower bound on the number of irrational quantities among the numbers $\zeta \left(3\right),\zeta \left(5\right),\dots ,\zeta \left(s\right)$.

We apply the theory of disconjugate or non-oscillatory three-, four-, and n-term linear recurrence relations on the real line to discuss some problems in number theory: specifically, to questions about the irrationality of various limits found as quotients of solutions at infinity and, in particular, to the irrationality and possible quadratic and higher algebraic irrationality of $\zeta \left(3\right)$ where $\zeta$ is the classic Riemann zeta function. We recall that this classic number is defined classically as

$$\zeta \left(3\right)=\sum _{n=1}^{\infty }\frac{1}{n^3}.$$

The motivation here is two-fold. First, we can investigate the irrationality of a given number L, say, by starting with an infinite series for L, associating it to a three-term recurrence relation (and so possibly to a non-regular continued fraction expansion) whose form is determined by the form of the series in question, finding an independent solution of said recurrence relation whose values are rational and, if the conditions are right (cf. Theorem 1 below), deduce the irrationality of L. We show that this abstract construction includes, as a special case, Apéry’s proof [1] that $\zeta \left(3\right)$ is irrational.

In trying to determine whether or not $\zeta \left(3\right)$ is an algebraic irrational [21], we address the question of whether $\zeta \left(3\right)$ is algebraic of degree two or more over $\mathbb{Q}$. Although we cannot answer this claim unequivocally at this time, we find an equivalent criterion for the irrationality of the square of $\zeta \left(3\right)$, or for that matter, any other irrational that can be approximated by the quotient of two solutions of an appropriate three-term recurrence relation. In the case of $\zeta \left(3\right)$, the equivalent criterion (Theorem 4) referred to is a function of the asymptotic behaviour of solutions of a specific linear four-term disconjugate recurrence relation (Theorem 2) in which the products of the classic Apéry numbers play a prominent role and whose general solution is actually known in advance. We obtain as a result that appropriate products of the Apéry numbers satisfy a four-term recurrence relation, that is, (32) below (indeed, given any $m\ge 2$, there exists an $(m+2)$-term recurrence relation for which these numbers play a basic role). However, the products of these Apéry numbers are not sufficient in themselves to give us the irrationality of the square of $\zeta \left(3\right)$. Still, our results show that the irrationality of the square of $\zeta \left(3\right)$ would imply the non-existence of linear combinations of appropriate products of Apéry sequences generating a principal solution of a certain type for this four-term linear recurrence relation. The converse is also true by our results, but we cannot show that such linear combinations do not exist. Hence, we cannot answer at this time whether $\zeta {\left(3\right)}^2$ is irrational.

We extend said criterion for the irrationality of a square of limits obtained by means of Apéry-type constructions, or from continued fraction expansions to a criterion for algebraic irrationality (an irrational satisfying a polynomial equation of degree greater than two with rational coefficients) over $\mathbb{Q}$ (Theorem 8). Here and everywhere else, $\mathbb{Q}$ denotes the set of all rational numbers. It is then a simple matter to formulate a criterion for the transcendence of such limits. Loosely speaking, we show that an irrational number derived as the limit of a sequence of rationals associated with a basis for a linear three-term recurrence relation is transcendental if and only if there exists an infinite sequence of linear m-term recurrence relations, one for each $m\ge 2$, such that each one lacks a non-trivial rational valued solution with special asymptotics at infinity (cf., Theorem 9). Finally, motivated by the results on the four-term recurrences (Theorem 2), we present in the Appendix A to this article accelerated series representations for $\zeta {\left(3\right)}^m$, for $m=2,3,4,5$, and similar series for $\zeta {\left(2\right)}^m$, where we display the cases $m=2,3$ only leaving the remaining cases as examples that can be formulated by the reader. These results are independent of those in [22] for ${\zeta }^2(2m+1)$ which use the generalized divisor function and Eisenstein series.

2. Preliminary Results

We present a series of lemmas useful in our later considerations. Recall that $\mathbb{R}$ is the set of all real numbers, $\mathbb{N}$ is the set of all non-negative integers (including zero), and ${\mathbb{Z}}^{+}$ is the set of all positive integers (excludes zero). Recall that ${\mathbb{Q}}^{+}$ denotes the set of all positive rational numbers.

Lemma 1. 

Let $A_n,c_n\in \mathbb{R}$, $n\in \mathbb{N}$, be two given infinite sequences such that the series

$$\sum _{n=1}^{\infty }\frac{1}{c_{n-1}{A}_n{A}_{n-1}}$$

(1)

converges absolutely. Then, there exists a sequence $B_n$ satisfying (5) such that

$$\underset{n\to \infty }{lim}\frac{B_n}{A_n}=\frac{B_0}{A_0}+\alpha \sum _{n=1}^{\infty }\frac{1}{c_{n-1}{A}_n{A}_{n-1}},$$

(2)

where $\alpha =c_0(A_0{B}_1-A_1{B}_0).$

Proof of Lemma 1. 

For the given sequences $A_n,c_n$ define the sequence $b_n$ using (3) below:

$$b_n=(c_n{A}_{n+1}+c_{n-1}{A}_{n-1})/A_n,n\ge 1.$$

(3)

Then, by definition, the $A_n$ satisfy the three-term recurrence relation

$$c_n{y}_{n+1}+c_{n-1}{y}_{n-1}-b_n{y}_n=0,n\in \mathbb{N},$$

(4)

with $y_0=A_0,y_1=A_1$. Choosing the values $B_0,B_1$ such that $\alpha \ne 0$, we solve the two-term recurrence relation

$$A_{n-1}{B}_n-A_n{B}_{n-1}=\frac{\alpha }{c_{n-1}},n\ge 1.$$

(5)

for a unique solution, $B_n$. Observe that these $B_n$ values satisfy the same recurrence relation as the given $A_n.$ Since $A_n{A}_{n-1}\ne 0$ by hypothesis, dividing both sides of (5) by $A_n{A}_{n-1}$ gives (2) upon summation and passage to the limit as $n\to \infty$, since the resulting series on the left is a telescoping series. □

Lemma 2. 

Consider (4) where$c_n>0$, $b_n-c_n-c_{n-1}>0$, for every$n\ge n_0\ge 1$, and${\sum }_{n=1}^{\infty }1/c_{n-1}<\infty .$Let$A_m,B_m\in \mathbb{R}$, $m\ge 1$, be two linearly independent solutions of (4). If$0\le A_0<A_1$, then

$$L\equiv \underset{m\to \infty }{lim}\frac{B_m}{A_m}$$

(6)

exists and is finite.

Proof of Lemma 2. 

Since $c_n>0$, $b_n-c_n-c_{n-1}>0$, for every $n\ge n_0$, Equation (4) is non-oscillatory at infinity [23] or [24]; that is, every solution $y_n$ has a constant sign for all sufficiently large n. From discrete Sturm theory, we deduce that every solution of (4) has a finite number of nodes [24]. As a result, the solution $A_n$, may, if modified by a constant, be assumed to be positive for all sufficiently large n. Similarly, we may assume that $B_n>0$ for all sufficiently large n. Thus, write $A_n>0,B_n>0$ for all $n\ge N.$ Once again, from standard results in the theory of three-term recurrence relations, there holds the Wronskian identity (5) for these solutions. The proof of Lemma 1, viz. (5), yields the identity

$$\frac{B_n}{A_n}-\frac{B_{n-1}}{A_{n-1}}=\frac{\alpha }{c_{n-1}{A}_n{A}_{n-1}},$$

(7)

for each $n\ge 1$. Summing both sides from $n=N+1$ to infinity, we deduce the existence of the limit L in (6) (possibly infinite at this point), since the tail end of the series has only positive terms and the left side is telescoping.

We now show that the eventually positive solution $A_n$ is bounded away from zero for all sufficiently large n. This is basically a simple argument (see Olver and Sookne [25] and Patula ([24], Lemma 2) for early extensions). Indeed, the assumption $0\le A_0<A_1$ implies that $A_n$ is increasing for all large n. An induction argument provides the clue. Assuming that $A_{k-1}\le A_k$ for all $1\le k\le n$,

$$A_{n+1}=(b_n{A}_n-c_{n-1}{A}_{n-1})/c_n\ge A_n(b_n-c_{n-1})/c_n>A_n,$$

since $b_n-c_n-c_{n-1}>0$ for all large n. The result follows.

Now since $A_n$ is bounded away from zero for large n (i.e., bounded below uniformly in n) and ${\sum }_{n=N+1}^{\infty }1/c_{n-1}<\infty$ by hypothesis, it follows that the series

$$\sum _{n=N+1}^{\infty }\frac{1}{c_{n-1}{A}_n{A}_{n-1}}<\infty ,$$

that is, L in (6) is finite. □

Remark 1. 

The limit of the sequence$A_n$itself may be a priori finite. For applications to irrationality proofs, we need that this sequence $A_n\to \infty$ with n. A sufficient condition for this is provided below.

Lemma 3. 

(Olver and Sookne [25], Patula ([24], Lemma 2)) Let$c_n>0$,

$$b_n-c_n-c_{n-1}>{\epsilon }_n{c}_n,$$

(8)

for all sufficiently large n, where${\epsilon }_n>0$, and${\sum }_{n=1}^{\infty }{\epsilon }_n$diverges. Then, every increasing solution$A_n$of (4) grows without bound as$n\to \infty$.

The notion of disconjugacy in its simplest form can be found in Patula [24] or see Hartman [23] for more general formulations. In our case, (4) is a disconjugate recurrence relation on $[0,\infty )$ if every non-trivial solution $y_n$ has at most one sign change for all $n\in \mathbb{N}$. The following result is a consequence of Lemmas 2 and 3.

Lemma 4. 

Let$c_n>0$in (4) and${\sum }_{n=1}^{\infty }1/c_{n-1}<\infty$. Let$b_n\in \mathbb{R}$satisfy

$$b_n-c_n-c_{n-1}>0,$$

for$n\ge 1$. Then

(a) 

Equation (4) is a disconjugate three-term recurrence relation on$[0,\infty ).$

(b) 

There exists a solution$A_n$with$A_n>0$for all$n\in \mathbb{N}$, $A_n$increasing and such that for any other linearly independent solution$B_n$, we have the relation

$$\left|L-\frac{B_m}{A_m}\right|\le \beta \frac{1}{A_m^2},$$

for some suitable constant β, for all sufficiently large m, where L is the limit.

(c) 

If, in addition, we have (8) satisfied for some sequence${\epsilon }_n>0$etc., then the solution$A_n$in item (2) grows without bound, that is, $A_n\to \infty$ as $n\to \infty$.

Item (b) of the preceding lemma is recognizable by anyone working with continued fractions [21]. Of course, continued fractions have convergents (such as $A_n,B_n$ above) that satisfy linear three-term recurrence relations and their quotients, when they converge, converge to the particular number (here represented by L) represented by the continued fraction. In this article, we view the limits of these quotients in terms of asymptotics of solutions of disconjugate recurrence relations with a particular emphasis on principal solutions.

3. Main Results

Theorem 1. 

Consider the three-term recurrence relation (4) where$b_n\in \mathbb{R}$, $c_n>0$for every $n\ge n_0\ge 1$, and the leading term$c_n$satisfies

$$\sum _{n=1}^{\infty }\frac{1}{c_{n-1}^k}<\infty ,$$

(9)

for some$k\ge 1$. In addition, let (8) be satisfied for some sequence${\epsilon }_n>0$, with${\sum }_{n=1}^{\infty }{\epsilon }_n=\infty$.

Let$0\le A_0<A_1$be given and the solution$A_m$of (4) satisfy$A_m\in {\mathbb{Q}}^{+}$for all large m, and for some δ,$0<\delta <k^{\prime }$and$k^{\prime }=k/(k-1)$whenever$k>1$, let

$$\sum _{n=1}^{\infty }\frac{1}{A_n^{\delta }}<\infty .$$

(10)

Next, let$B_m$be a linearly independent solution such that$B_m\in \mathbb{Q}$for all sufficiently large m and such that for some sequences$d_m,e_m\in {\mathbb{Z}}^{+}$, we have$d_m{A}_m\in {\mathbb{Z}}^{+}$and$e_m{B}_m\in {\mathbb{Z}}^{+}$, for all sufficiently large m, and

$$\underset{m\to \infty }{lim}\frac{\mathrm{lcm}(d_m,e_m)}{{A_m}^{1-\delta /k^{\prime }}}=0.$$

(11)

Then, L, defined in (6), is an irrational number.

Proof of Theorem 1. 

We separate the cases $k=1$ from $k>1$ as is usual in this kind of argument. Let $k=1$. With $A_n,B_n$ as defined, a simple application of Lemma 2 (see (7)) gives us that for $m\ge N$,

$$\sum _{n=m+1}^{\infty }\left(\frac{B_n}{A_n}-\frac{B_{n-1}}{A_{n-1}}\right)=\alpha \sum _{n=m+1}^{\infty }\frac{1}{c_{n-1}{A}_n{A}_{n-1}},$$

(12)

i.e.,

$$L-\frac{B_m}{A_m}=\alpha \sum _{n=m+1}^{\infty }\frac{1}{c_{n-1}{A}_n{A}_{n-1}}.$$

(13)

Since $A_n$ is increasing for all $n\ge N$ (by Lemma 2), we have $A_n{A}_{n-1}>A_{n-1}^2$ for such n. In fact, we also have $A_n\to \infty$ (by Lemma 3). Estimating (13) in this way, we obtain

$$\left|L-\frac{B_m}{A_m}\right|\le \alpha \sum _{n=m+1}^{\infty }\frac{1}{c_{n-1}{A}_{n-1}^2},$$

(14)

and since $A_k^2>A_m^2$ for $k>m$, we obtain

$$\left|L-\frac{B_m}{A_m}\right|\le \beta \frac{1}{A_m^2},$$

(15)

where $\beta =\alpha {\sum }_{n=m+1}^{\infty }1/c_{n-1}<\infty$. Multiplying (15) by $\mathrm{lcm}(d_m,e_m)·A_m$ for all large m, we find

$$\left|\mathrm{lcm}(d_m,e_m)A_{m}L-B_m\mathrm{lcm}(d_m,e_m)\right|\le \beta \frac{\mathrm{lcm}(d_m,e_m)}{A_m}.$$

(16)

Assuming that $L=C/D$ is rational, where $C,D$ are relatively prime, we obtain

$$\left|\mathrm{lcm}(d_m,e_m)A_{m}C-B_{m}D\mathrm{lcm}(d_m,e_m)\right|\le \beta D\frac{\mathrm{lcm}(d_m,e_m)}{A_m}.$$

However, the left-hand side is a non-zero integer for every m (see (13)), while the right side tends to zero as $m\to \infty$ by (11) with $k^{\prime }=\infty$. Hence, it must eventually be less than 1, for all large m, and this leads to a contradiction. This completes the proof for $k=1$.

Let $k>1$. Proceed as in the case $k=1$ up to (14). The proof of Lemma 2 shows that the solution $A_n$ as defined is increasing. The fact that this $A_n\to \infty$ as $n\to \infty$ follows from Lemma 3. The existence of the limit L is clear, since the series consists of positive terms for all sufficiently large m. In order to prove that this L is finite, observe that

$$\left|L-\frac{B_m}{A_m}\right|\le \beta {\left(\sum _{n=m+1}^{\infty }\frac{1}{A_n^{k^{\prime }}{A}_{n-1}^{k^{\prime }}}\right)}^{1/k^{\prime }}$$

where $\beta =\alpha {\left({\sum }_{n=m+1}^{\infty }1/c_{n-1}^k\right)}^{1/k}<\infty$, by (9). Next, $A_n^{k^{\prime }}{A}_{n-1}^{k^{\prime }}$$=A_n^{\delta }{A}_n^{k^{\prime }-\delta }{A}_{n-1}^{k^{\prime }}$$\ge A_n^{\delta }{A}_m^{2k^{\prime }-\delta }$, for all large n. Hence,

$$\left|L-\frac{B_m}{A_m}\right|\le \frac{{\beta }^{\prime }}{A_m^{2-\delta /k^{\prime }}},$$

(17)

where ${\beta }^{\prime }=\beta {({\sum }_{n=m+1}^{\infty }1/A_n^{\delta })}^{1/k^{\prime }}<\infty$ by (10). Since $0<\delta <k^{\prime }$, we learn that L is finite. Equation (17) corresponds to (15) above. Continuing as in the case $k=1$ with minor changes, we see that (11) is sufficient for the irrationality of L. □

Remark 2. 

In the case$k=1$, condition (10) is not needed. This condition is verified for corresponding solutions of (4) with$c_n=n+1$, $b_n=an+b$where$a>2$, for all sufficiently large indices. Note that$a=2$is a borderline case. For example, for$a=2,b=1$, there are both bounded non-oscillatory solutions (e.g.,$y_n=1$) and unbounded non-oscillatory solutions (e.g.,$y_n=1+3\psi (n+1)+3\gamma$, where$\psi \left(x\right)={(log\Gamma \left(x\right))}^{\prime }$is the digamma function and γ is Euler’s constant). So, for every pair of such solutions, the limit L is either infinite or rational. For $a<2$ all solutions are oscillatory, that is $y_n{y}_{n-1}<0$ for all arbitrarily large indices. Such oscillatory cases could also be of interest for number theoretical questions, especially so if the ratio of two independent solutions is of one sign for all sufficiently large n (as in Zudilin [14]).

3.1. Consequences

One of the simplest consequences involves yet another interpretation of the proof of the irrationality of $\zeta \left(3\right)$ and of $\zeta \left(2\right)$. It mimics many of the known proofs, yet a large part of it involves only the theory of disconjugate three-term recurrence relations as studied here. As the proofs are similar, we will sketch the proof only for the case of $\zeta \left(3\right)$.

Proposition 1. 

$\zeta \left(3\right)$is irrational.

Proof of Proposition 1. 

(Apéry [1], see also Van der Poorten [26], Beukers [2], and Cohen [27]).

Consider (4) with $c_n={(n+1)}^3$ and $b_n=34n^3+51n^2+27n+5$, $n\ge 0$. This gives Apéry’s recurrence relation,

$${(n+1)}^3{y}_{n+1}+n^3{y}_{n-1}=(34n^3+51n^2+27n+5)y_n,n\ge 1.$$

(18)

Define two independent solutions $A_n,B_n$ of (18) by the initial conditions $A_0=1,A_1=5$ and $B_0=0,B_1=6$. Then, $b_n-c_n-c_{n-1}>0$ for every $n\ge 0$. Since $0<A_0<A_1$ by Lemma 2, the sequence $A_n$ is increasing and tends to infinity as $n\to \infty$. In addition to $c_n>0$, (8) is satisfied for every $n\ge 1$ and ${\epsilon }_n=1/n$, say. Hence, (18) is a disconjugate three-term recurrence relation on $[0,\infty )$. An application of Lemmas 2 and 3 shows that

$$L\equiv \underset{m\to \infty }{lim}\frac{B_m}{A_m}$$

(19)

exists and is finite and, as a by-product, we obtain (2), that is (since $B_0=0$),

$$L=\alpha \sum _{n=1}^{\infty }\frac{1}{c_{n-1}{A}_n{A}_{n-1}},$$

(20)

where $\alpha =6$ in this case.

Define non-negative sequences $A_n,B_n$ by setting

$$A_n=\sum _{k=0}^n{\left(\genfrac{}{}{0pt}{}{n}{k}\right)}^2{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}^2,$$

(21)

and

$$B_n=\sum _{k=0}^n{\left(\genfrac{}{}{0pt}{}{n}{k}\right)}^2{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}^2\left(\sum _{m=1}^n\frac{1}{m^3}+\sum _{m=1}^k\frac{{(-1)}^{m-1}}{2m^3\left(\genfrac{}{}{0pt}{}{n}{m}\right)\left(\genfrac{}{}{0pt}{}{n+m}{m}\right)}\right).$$

(22)

A long and tedious calculation (see Cohen [27]) gives that these sequences satisfy (18) and thus must agree with our solutions (bearing the same name), since their initial values agree. That $L=\zeta \left(3\right)$ in (19) is shown directly by using these expressions for $A_n,B_n$. In addition, it is clear that $A_n\in {\mathbb{Z}}^{+}$ (so $d_n=1$ in Theorem 1) while the $B_n\in {\mathbb{Q}}^{+}$ have the property that if $e_m=2\mathrm{lcm}{[1,2,\dots n]}^3$, then $e_m{B}_m\in {\mathbb{Z}}^{+}$, for every $m\ge 1$ (cf., e.g., [26,27] among many other such proofs). Hence, the remaining conditions of Theorem 1 are satisfied, for $k=1$ there. So, since it is known that asymptotically $e_m/A_m\to 0$ as $m\to \infty$ (e.g., [26]), the result follows from said theorem. □

Remark 3. 

Strictly speaking, the number thoeretical part only comes into play after (20). If we knew somehow that the series in (20) summed to$\zeta \left(3\right)$independently of the relations (21) and (22) that follow, we would have a more natural proof. This is not a simpler proof of the irrationality of$\zeta \left(3\right)$; it is simply a restatement of the result in terms of the general theory of recurrence relations in yet another approach to the problem of irrationality proofs. The proof presented is basically a modification of Cohen’s argument in [27] recast as a result in the asymptotic theory of three-term recurrence relations. We also observe that a consequence of the proof is that ([Fischler [4], Remarque 1.3, p. 910-04]),

$$\zeta \left(3\right)=6\sum _{n=1}^{\infty }\frac{1}{n^3{A}_n{A}_{n-1}},$$

(23)

an infinite series that converges much faster (series acceleration) to$\zeta \left(3\right)$than the original series considered by Apéry, that is

$$\zeta \left(3\right)=\frac{5}{2}\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^3\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}.$$

(24)

For example, the first five terms of the series (23) gives 18 correct decimal places to$\zeta \left(3\right)$while (24) only gives four. At the end of this paper, we provide some series acceleration for arbitrary integral powers of$\zeta \left(3\right)$.

The preceding remark leads to the following natural scenario. Let us say that we start with the infinite series

$$L=\sum _{n=1}^{\infty }\frac{1}{n^3{A}_n{A}_{n-1}}$$

(25)

where the terms $A_n$ are the Apéry numbers defined in (21) and the series (25) has been shown to be convergent using direct means, that is, avoiding the use of the recurrence relation (18). Then, by Lemma 1, there exists a rational valued sequence $B_n$ such that both $A_n,B_n$ are linearly independent solutions of a three-term recurrence relation of the form (4). The new sequence $B_n$ thus obtained must be a constant multiple of their original counterpart in (22). Solving for the $b_n$ using (2) would necessarily give the cubic polynomial in (18), which has since been a mystery. Once we have the actual recurrence relation in question, we can then attempt an irrationality proof of the number L using the methods described, the only impediment being how to show that $e_m{B}_m\in {\mathbb{Z}}^{+}$ without having an explicit formula such as (22).

The method can be summed up generally as follows: We start with an infinite series of the form

$$L=\sum _{n=1}^{\infty }\frac{1}{c_{n-1}{A}_n{A}_{n-1}}$$

(26)

where the terms $c_n,A_n\in {\mathbb{Z}}^{+}$, and the series (26) has been shown to be convergent to L using some direct means. Then, by Lemma 1, there exists a rational valued sequence $B_n$ such that both $A_n,B_n$ are linearly independent solutions of (4) where the $b_n$, defined by (2), are rational for every n. If, in addition, we have for example,

$$\sum _{n=1}^{\infty }1/c_{n-1}<\infty ,$$

(27)

along with (8), we can then hope to be in a position so as to apply Theorem 1 and obtain the irrationality of the real number L. Of course, this all depends on the interplay between the growth of the $d_n{A}_n$ at infinity and the rate of growth of the sequence $e_n{B}_n$ required by said Theorem (see (11)). The point is that the relation (15) used by some to obtain irrationality proofs for the number L is actually a consequence of the theory of disconjugate three-term recurrence relations. In fact, underlying all this is Lemma 4.

The next two results are expected and included because their proofs are instructive for later use.

Proposition 2. 

The only solution of (18) whose values are all positive integers is, up to a constant multiple, the solution$A_n$in (21).

Proof of Proposition 2. 

If possible, let $D_n$ be another integer valued solution of (18). Then, $D_n=a{A}_n+b{U}_n$, for every $n\in \mathbb{N}$ where $a,b\in \mathbb{R}$ are constants. Using the initial values $A_0=1,A_1=5$, $U_0=\zeta \left(3\right),U_1=5\zeta \left(3\right)-6$, in the definition of $D_n$, we deduce that $a=D_0-(5D_0-D_1)\zeta \left(3\right)/6$ and $b=(5D_0-D_1)/6.$ Thus,

$$D_n=D_0{A}_n-(5D_0-D_1)B_n/6,n\ge 1,$$

where the coefficients of $A_n,B_n$ above are rational numbers. By hypothesis, the sequence $D_n$, $n\in \mathbb{N}$ is integer valued. However, so is $A_n$; thus, $D_n-D_0{A}_n\in \mathbb{Z}$ for all n. Therefore, for $5D_0-D_1\ne 0$, we must have that $(5D_0-D_1)B_n/6\in \mathbb{Z}$ for all n, which is impossible for sufficiently large n (see (22)). Hence, $5D_0-D_1=0$, and this shows that $D_n$ must be a multiple of $A_n$. □

Proposition 3. 

The solution$B_n$of Ap is not unique. That is, there exists an independent strictly rational (i.e., non-integral) solution$D_n$of (18) such that

$$\frac{1}{3}\mathrm{lcm}{[1,2,\dots ,n]}^3{D}_n\in {\mathbb{Z}}^{+}$$

for all n.

Proof of Proposition 3. 

A careful examination of the proof of Proposition 2 shows that the solution $B_n$ defined in (22) is not the only solution of (18) with the property that $2\mathrm{lcm}{[1,2,\dots ,n]}^3{B}_n\in {\mathbb{Z}}^{+}$ for all n. Indeed, the solution $D_n$, defined by setting $D_0=1$, $D_1=1$ and $D_n=D_0{A}_n-(5D_0-D_1)B_n/6$, for $n\ge 1,$ has the additional property that $2\mathrm{lcm}{[1,2,\dots ,n]}^3{D}_n/6\in {\mathbb{Z}}^{+}$ for all n. Thus, the claim is that $2\mathrm{lcm}{[1,2,\dots ,n]}^3{B}_n$ is divisible by 6 for every $n\in \mathbb{N}$. So, it suffices to show that $\mathrm{lcm}{[1,2,\dots ,n]}^3{B}_n$ is divisible by 3. This can be shown by considering the contribution of this divisor to the p-adic valuation, $v_p(·)$, of one term in the third sum of (22). Look at Cohen’s proof ([27], Proposition 3) where it is shown that $2\mathrm{lcm}{[1,2,\dots ,n]}^3{B}_n\in {\mathbb{Z}}^{+}$ for all n. There, he shows that

$$v=v_p\left(\frac{d_n^3\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}{m^3\left(\genfrac{}{}{0pt}{}{n}{m}\right)\left(\genfrac{}{}{0pt}{}{n+m}{m}\right)}\right)\ge \dots \ge (v_p\left(d_n\right)-v_p\left(m\right))+(v_p\left(d_n\right)-v_p\left(d_k\right))\ge 0,$$

where $d_n=\mathrm{lcm}{[1,2,\dots ,n]}^3.$ Observe that insertion of the factor $1/3$ on the left only decreases the right side by 1 for the 3-adic valuation (see [27], p.VI.5) and then, keeping track of the other two terms above on the right and the fact that they are not zero, we see that the inequality is still valid. Of course, one cannot do better than the divisor ‘6’ in this respect since $B_1=6$. □

Remark 4. 

A simple heuristic argument in the case of$\zeta \left(5\right)$shows that if we are looking for recurrence relations of the form (4) with$c_n={(n+1)}^5$and$b_n$some quintic polynomial in n, and we want an integral-valued solution other than the trivial ones, then we must have the coefficient of the leading term of the quintic superior to 150 in order for the asymptotics to work out at all. The subsequent existence of a second solution$S_n$with the property that$c·lcm{[1,2,\dots ,n]}^5{S}_n\in {\mathbb{Z}}^{+}$for all n, where c is a universal constant, is then not out of the question and could lead to an irrationality proof of this number. However, it is not at all clear to us that such a (non-trivial) quintic exists.

The basic advantage of the formalism of recurrence relations lies in that every element in $\mathbb{Q}\left(\zeta \right(3\left)\right)$ can be approximated by ‘good’ rationals, that is appropriate linear combinations of the $A_n,B_n$ in (21) and (22). For example, the series considered by Wilf [28]

$$\sum _{n=1}^{\infty }\frac{1}{n^3{(n+1)}^3{(n+2)}^3}=\frac{29}{32}-\frac{3}{4}\zeta \left(3\right),$$

derived as a result of the use of the WZ algorithm, e.g., [29], has a counterpart via (18). The solution $C_n$ of (18) defined by $C_n=(29/32)A_n-(3/4)B_n$ has the property that $C_n/A_n\to (29/32)-(3/4)\zeta \left(3\right)$ as $n\to \infty$, and the convergence of these fractions is sufficiently rapid as to ensure the irrationality of its limit, but this does not appear to be so for Wilf’s series, even though it is an ‘accelerated’ series. A similar comment applies to the series

$$\sum _{m=1}^{\infty }\frac{1}{{(m+1)}^3{(m+2)}^3{(m+3)}^3{(m+4)}^3{(m+5)}^3}=\frac{5}{768}\zeta \left(3\right)-\frac{10385}{98304},$$

also derived in [28]. We point out that the above two series can also be summed more simply by using the method of partial fractions.

The usefulness of so-called dominant and recessive solutions in the theory (also called principal solutions by some) is apparent in the following discussion regarding the overall nature of the solutions of (18). As noted earlier, $A_n>0$ for every n, $A_n$ is increasing, and the series in (2) converges. In addition, by defining the solution $U_n=\zeta \left(3\right)A_n-B_n$, we see that $U_n/A_n\to 0$ as $n\to \infty$ (see the proof of Proposition 1). Hence, by definition, $A_n$ (resp. $U_n$) is a dominant (resp. recessive) solution of the disconjugate Equation (18) on $[0,\infty )$, and as a dominant (resp. recessive) solution, it is unique up to a constant multiple, [23,24].

In this paragraph, we fix a pair of dominant/recessive solutions of (18), say, $A_n$ and $U_n$, respectively. Let $L>0$. Then, there is a sequence of reals of the form $V_n/A_n$, where $V_n$ is a solution of (18) such that $V_n/A_n\to L$ as $n\to \infty$. Indeed, choose $V_n$ by setting $V_n=U_n+L{A}_n$. Hence, for example, there exists a solution $V_n$ of (18) such that

$$V_n/A_n\to \zeta \left(5\right)n\to \infty ,$$

or another solution $W_n$ such that

$$W_n/A_n\to \zeta \left(7\right)n\to \infty ,\mathrm{etc}.$$

but the terms of $V_n,W_n$, etc. are not necessarily all rational. In addition, for a given real $L>0$ and any $\gamma >0$, the solution $V_n\equiv B_n+\gamma U_n$ is such that $V_n/A_n\to \zeta \left(3\right),$ as $n\to \infty$.

3.2. On the Irrationality of $\zeta {\left(3\right)}^2$

Another question is whether $\zeta \left(3\right)$ is itself algebraic of degree 2 over $\mathbb{Q}$? Although we do not answer this question either way, we present an apparently tractable equivalent formulation which may shed some light on this question. The method is sufficiently general so as to show that given any number known to be irrational by applying an Apéry-type argument on a three-term recurrence relation or issuing from a continued fraction expansion, the statement that it is a number whose square is irrational is equivalent to a statement about rational valued principal solutions of a corresponding disconjugate four-term recurrence relation.

We proceed first by showing that solutions of a linear three-term recurrence relation can be used to generate a basis for a corresponding four-term linear recurrence relation. The analogous result for differential equations is sufficiently well-known and old; see e.g., Ince [30]. Our corresponding result, Theorem 2 below, appears to be new in the general case. As a consequence, the quantities $A_n,B_n$ defined in (21) and (22) can be used to generate a basis for a new recurrence relation of order one higher than the original one (18) considered by Apéry.

Given any three-term recurrence relation in general form

$$p_n{y}_{n+1}+q_n{y}_{n-1}=r_n{y}_n,n\ge 1,$$

(28)

the mere assumption that $p_n{q}_n\ne 0$ for all n enables one to transform (28) into the self-adjoint form (29) below by means of the substitution $c_n=c_{n-1}{p}_n/q_n$, $c_0$ given, and $b_n=c_n{r}_n/p_n$. Hence, for simplicity and ease of exposition, we assume that the recurrence relation is already in self-adjoint form, and there is no loss of generality in assuming this. We maintain the use of the symbols $A_n,B_n$ for the solutions under consideration for motivational purposes.

Theorem 2. 

Let$A_n,B_n$generate a basis for the solution space of the three-term recurrence relation (29)

$$c_n{y}_{n+1}+c_{n-1}{y}_{n-1}-b_n{y}_n=0,n\ge 1,$$

(29)

where$c_n\ne 0$, $b_n\ne 0$for every n, and$b_n,c_n\in \mathbb{R}$. Then, the quantities $x_{n-1}\equiv A_n{A}_{n-1}$, $y_{n-1}\equiv B_n{B}_{n-1}$, $z_{n-1}\equiv A_n{B}_{n-1}+A_{n-1}{B}_n$form a basis for the space of solutions of the four-term recurrence relation

$$\begin{array}{ccc}\hfill & & c_{n+2}{c}_{n+1}^2{b}_n{z}_{n+2}+(b_n{c}_{n+1}^3-b_n{b}_{n+1}{b}_{n+2}{c}_{n+1})z_{n+1}+\hfill \\ & & (b_n{b}_{n+1}{b}_{n+2}{c}_n-b_{n+2}{c}_n^3)z_n-c_{n-1}{c}_n^2{b}_{n+2}{z}_{n-1}=0,n\ge 1,\hfill \end{array}$$

(30)

Proof of Theorem 2. 

Direct verification using repeated applications of (29) and simplification, we omit the details. The linear independence can be proved using Wronskians; see below (and see Hartman [23] but where in Proposition 2.7 on p. 8 of [23], the reader should replace a by $\alpha$). □

The Wronskian of the three solutions $x_n=A_{n+1}{A}_n$, $y_n=B_{n+1}{B}_n$, $z_n=A_{n+1}{B}_n+A_n{B}_{n+1}$ of (30) arising from the two independent solutions $A_n,B_n$ of the three-term recurrence relation (29) is given by the determinant of the matrix ([31], p. 310),

$$\left[\begin{array}{ccc}{A}_{n+1}{A}_n& B_{n+1}{B}_n& A_{n+1}{B}_n+A_n{B}_{n+1}\\ A_{n+2}{A}_{n+1}& B_{n+2}{B}_{n+1}& A_{n+2}{B}_{n+1}+A_{n+1}{B}_{n+2}\\ A_{n+3}{A}_{n+2}& B_{n+3}{B}_{n+2}& A_{n+3}{B}_{n+2}+A_{n+2}{B}_{n+3}\end{array}\right]$$

which, after the usual iterations (or see [[23], Prop.2.7]) reduces to the expression:

$$\frac{b_{n+2}{b}_{n+1}{c}_{n-1}^3{\left(A_n{B}_{n-1}-B_n{A}_{n-1}\right)}^3}{c_n{c}_{n+2}{c}_{n+1}^3}.$$

(31)

We apply Theorem 2 to the questions at hand, although it is likely there are more numerous applications elsewhere. Thus, the following corollary (stated as a theorem) is immediate.

Theorem 3. 

Let$A_n,B_n$be the Apéry sequences define above in (21) and (22) and consider the corresponding three-term recurrence relation (18) where, for our purposes,$c_n={(n+1)}^3,b_n=34n^3+51n^2+27n+5$. Then, the four-term recurrence relation

$$\begin{array}{ccc}\hfill & & {\left(n+3\right)}^3{\left(n+2\right)}^6\left(2n+1\right)\left(17n^2+17n+5\right)z_{n+2}\hfill \\ & & -\left(2n+1\right)\left(17n^2+17n+5\right)(1155n^6+13860n^5+68535n^4\hfill \\ & & +178680n^3+259059n^2+198156n+62531){\left(n+2\right)}^3{z}_{n+1}\hfill \\ & & +\left(2n+5\right)\left(17n^2+85n+107\right)(1155n^6+6930n^5+16560n^4\hfill \\ & & +20040n^3+12954n^2+4308n+584){\left(n+1\right)}^3{z}_n\hfill \\ & & -{\left(n+1\right)}^6{n}^3\left(2n+5\right)\left(17n^2+85n+107\right)z_{n-1}=0,\hfill \end{array}$$

(32)

admits each of the three products$x_{n-1}\equiv A_n{A}_{n-1}$, $y_{n-1}\equiv B_n{B}_{n-1}$, and$z_{n-1}\equiv A_n{B}_{n-1}+A_{n-1}{B}_n$as a solution, and the resulting set is a basis for the solution space of (32).

The calculation of the Wronskian in the case of (32) is an now easy matter (see (31)). In the case of our three solutions of (32), namely $x_n,y_n,z_n$ defined in Theorem 2, the Wronskian is given by

$$\frac{\left(2n+3\right)\left(2n+5\right)\left(17n^2+51n+39\right)\left(17n^2+85n+107\right)n^9{\left(A_n{B}_{n-1}-B_n{A}_{n-1}\right)}^3}{{\left(n+1\right)}^3{\left(n+2\right)}^9{\left(n+3\right)}^3}$$

The non-vanishing of the determinant for every n is also clear. The counterpart to (5) in this higher-order setting is

$$R_n{\left(A_n{B}_{n-1}-B_n{A}_{n-1}\right)}^3=L_{n}detW(x,y,z)\left(0\right),$$

where $W(x,y,z)\left(0\right)=-62595/64$,

$$R_n\equiv \frac{\left(2n+3\right)\left(2n+5\right)\left(17n^2+51n+39\right)\left(17n^2+85n+107\right)n^9}{{\left(n+1\right)}^3{\left(n+2\right)}^9{\left(n+3\right)}^3},$$

and

$$L_n\equiv \prod _{m=1}^n\frac{m^3{(m+1)}^6(2m+5)(17m^2+85m+107)}{{(m+2)}^6{(m+3)}^3(2m+1)(17m^2+17m+5)}.$$

We recall the fact that $A_n{A}_{n-1}$ is a solution of (32), that $A_0=1$, $A_1=5$, and $A_n>0$ for every $n>1$.

Theorem 4. 

$\zeta \left(3\right)$is algebraic of degree two over$\mathbb{Q}$if and only if (32) has a non-trivial rational valued solution$S_n$(i.e.,$S_n$is rational for every$n\ge 1$), with

$$\frac{S_n}{A_n{A}_{n-1}}\to 0,n\to \infty .$$

(33)

Proof of Theorem 4. 

(Sufficiency) Since $A_n{A}_{n-1}$, $B_n{B}_{n-1}$ and $A_{n-1}{B}_n+A_n{B}_{n-1}$ are linearly independent, we have

$$S_n=a{A}_n{A}_{n-1}+b{B}_n{B}_{n-1}+c(A_{n-1}{B}_n+A_n{B}_{n-1}),$$

(34)

for some $a,b,c\in \mathbb{R}$, not all zero. Since $S_n$ is rational valued for all n by hypothesis, the repeated substitutions $n=1,2,3$ in the above display yield a system of three equations in the unknowns $a,b,c.$ Since all the values involved are rational numbers, the same is true of this unique set of $a,b,c$.

With this set of $a,b,c$ we note that for every $n\ge 1$,

$$\frac{S_n}{A_n{A}_{n-1}}=a+b\frac{B_n{B}_{n-1}}{A_n{A}_{n-1}}+c\left(\frac{B_n}{A_n}+\frac{B_{n-1}}{A_{n-1}}\right).$$

(35)

However, from Apéry’s work [1] (or [6,26]), we know that $B_n/A_n\to \zeta \left(3\right)$ as $n\to \infty$. Using this information in passing to the limit, we have that

$$\frac{S_n}{A_n{A}_{n-1}}\to a+b\zeta {\left(3\right)}^2+2c\zeta \left(3\right),n\to \infty .$$

(36)

The possibility that $b=0$ is excluded by the fact that $\zeta \left(3\right)$ is irrational. Thus, $b\ne 0$ and so $\zeta \left(3\right)$ is algebraic of degree 2 over $\mathbb{Q}$.

Conversely, assume that $\zeta \left(3\right)$ is algebraic of degree 2 over $\mathbb{Q}$. Then, there exists rational constants $a,b,c$ with $b\ne 0$ such that $b\zeta {\left(3\right)}^2+2c\zeta \left(3\right)+a=0.$ For this choice of $a,b,c,$ consider the solution of (32) defined by (34). Since this $S_n$ clearly satisfies (35), and (36) by construction, the limiting result (33) follows. □

The proof of Theorem 4 is capable of much greater generality. Combined with Theorem 2 and minor changes in the argument of the previous theorem, one can easily prove

Theorem 5. 

Let$c_n,b_n\in \mathbb{R}$, $c_n\ne 0$, $b_n\ne 0$, for every n. Let$A_n,B_n$be two independent rational valued solutions of (29) such that

$$\underset{n\to \infty }{lim}\frac{B_n}{A_n}=L,$$

where L is irrational. Then, L is algebraic of degree two over$\mathbb{Q}$if and only if (30) has a non-trivial rational valued solution$S_n$such that (33) holds.

Example 1. 

Consider the Fibonacci sequence$F_n$defined by the self-adjoint three-term recurrence relation (29) with$b_n=c_n={(-1)}^n$for all$n\ge 1$, and$c_0=1.$Then, it is easy to see that the solutions defined by$A_0=0$, $A_1=1$, $B_0=1$and$B_1=1$are given by$A_n\equiv F_n$, $B_n=F_{n+1}$are two linearly independent solutions of the Fibonacci relation such that$B_n/A_n\to L$, where$L=(1+\sqrt{5})/2$is already known to be irrational (it is not necessary that L be irrational as a result of the actual approach to the limit).

According to Theorem 2, the quantities $F_n{F}_{n-1}$, $F_{n+1}{F}_n$ and $F_n^2+F_{n-1}{F}_{n+1}$ satisfy the four-term recurrence relation

$$z_{n+2}-2z_{n+1}-2z_n+z_{n-1}=0.$$

Note that the solution $S_n$ defined by

$$S_n=-F_n{F}_{n-1}+F_{n+1}{F}_n-(F_n^2+F_{n-1}{F}_{n+1})/2,$$

is a non-trivial rational valued solution of this four term recurrence relation such that $S_n/F_n{F}_{n-1}\to L^2-L-1=0$ as $n\to \infty$. Hence, by Theorem 2, L is algebraic of degree 2.

3.3. Discussion

In the language of the theory of disconjugate difference equations, a special solution such as $S_n$ in Theorem 4, if it exists, is a 2nd principal solution of (32). In the case of a disconjugate four-term linear recurrence relation with a positive leading term (such as ours, (32)), a $k\mathrm{th}$ principal solution $u_{k,n}$ is characterized by the existence of limits of the form

$$\frac{u_{k-1,n}}{u_{k,n}}\to 0,n\to \infty ,$$

for $k=1,2,3$. A first principal solution $u_{0,n}$ is unique up to a multiplicative constant, when it exists. For example, in the case of $\zeta \left(3\right)$, (7) gives us that quotients of solutions of disconjugate linear recurrence relations always have limits at infinity (and they are allowed to be infinite). In the case of (32), this is easy to see, since we know the basis explicitly. For example, the limit of the two solutions $u_n=B_n{B}_{n-1}$ and $v_n=2\zeta \left(3\right)A_n{A}_{n-1}-(A_n{B}_{n-1}+A_{n-1}{B}_n)$ of (32) exists at infinity, and ${lim}_{n\to \infty }{u}_n/v_n=+\infty$. On the other hand, the solution $v_n$ just defined and $w_n=A_n{A}_{n-1}+A_n{B}_{n-1}+A_{n-1}{B}_n$ are such that ${lim}_{n\to \infty }{v}_n/w_n=0.$ In the case of disconjugate or non-oscillatory difference equations (or recurrence relations), such principal solutions always exist, see Hartman ([23], Section 8), ([31], Appendix A) for basic discussions on these and related matters.

4. A Criterion for Algebraic Irrationality and Transcendence

In this section, we generalize the results of the previous sections. Thus, given an irrational number L whose rational approximations are found by either an Apéry-type argument on a three-term recurrence relation, or perhaps using a continued fraction expansion of L, we prove that if L is not algebraic of degree less than or equal to $m-1$, then L is algebraic of degree m (over $\mathbb{Q}$) if and only if there exists a disconjugate $(m+2)$-term linear recurrence relation having a specific type of rational valued principal solution. This will lead to Theorem 5 above as a special case. This will lead to, of course, a necessary and sufficient condition for the transcendence of such quantities. We outline the construction of this characteristic recurrence relation by pointing out the first two important special cases first as motivation: The first case is Theorem 5 as mentioned above. The second case is a “degree 3” version of Theorem 5.

In relation to (29) can be found a higher order analog of Theorem 2. We start with the five-term recurrence relation

$$p_n{z}_{n+3}-q_n{z}_{n+2}-r_n{z}_{n+1}-s_n{z}_n-t_n{z}_{n-1}=0,n\ge 1,$$

(37)

where the leading term $p_n\ne 0$ for all n, and

$$\begin{array}{c}{p}_n=c_{n+4}{c_{n+3}}^2{c_{n+2}}^3{b}_{n+1}\left({c_n}^2-b_{n+1}{b}_n\right),\hfill \\ q_n=-b_{n+2}{b}_{n+1}{c_{n+3}}^3{c_{n+2}}^2{c_n}^2+{b_{n+1}}^2{c}_{n+3}{c_{n+2}}^4{b}_{n+4}{b}_n-\hfill \\ b_{n+2}{b_{n+1}}^2{b}_{n+3}{c}_{n+3}{c_{n+2}}^2{b}_{n+4}{b}_n+b_{n+2}{b}_{n+1}{b}_{n+3}{c}_{n+3}{c_{n+2}}^2{b}_{n+4}{c_n}^2-\hfill \\ b_{n+1}{c}_{n+3}{c_{n+2}}^4{b}_{n+4}{c_n}^2+b_{n+2}{b_{n+1}}^2{c_{n+3}}^3{c_{n+2}}^2{b}_n\hfill \\ \\ r_n=-b_{n+2}{b}_{n+1}{b_{n+3}}^2{c}_{n+2}{c}_{n+1}{b}_{n+4}{c_n}^2-{b_{n+1}}^2{c_{n+2}}^3{c}_{n+1}{b}_{n+4}{b}_n{b}_{n+3}+\hfill \\ b_{n+1}{b}_{n+3}{c}_{n+2}{c_{n+1}}^3{c_{n+3}}^2{b}_n+b_{n+2}{b}_{n+1}{b}_{n+3}{c}_{n+2}{c}_{n+1}{c_{n+3}}^2{c_n}^2-\hfill \\ b_{n+1}{c}_{n+2}{c_{n+1}}^3{b}_{n+4}{b}_n{b_{n+3}}^2+{b_{n+1}}^2{c_{n+2}}^3{c}_{n+1}{c_{n+3}}^2{b}_n+\hfill \\ b_{n+1}{b}_{n+3}{c_{n+2}}^3{c}_{n+1}{b}_{n+4}{c_n}^2+{b_{n+3}}^2{c}_{n+2}{c_{n+1}}^3{b}_{n+4}{c_n}^2-\hfill \\ b_{n+1}{c_{n+2}}^3{c}_{n+1}{c_{n+3}}^2{c_n}^2-b_{n+2}{b_{n+1}}^2{b}_{n+3}{c}_{n+2}{c}_{n+1}{c_{n+3}}^2{b}_n+\hfill \\ b_{n+2}{b_{n+1}}^2{c}_{n+2}{c}_{n+1}{b}_{n+4}{b}_n{b_{n+3}}^2-b_{n+3}{c}_{n+2}{c_{n+1}}^3{c_{n+3}}^2{c_n}^2,\hfill \\ \\ s_n=-b_{n+3}{c_{n+1}}^4{c}_n{c_{n+3}}^2{b}_n+b_{n+2}{c_{n+1}}^2{c_n}^3{b}_{n+4}{b_{n+3}}^2+\hfill \\ {b_{n+3}}^2{c_{n+1}}^4{c}_n{b}_{n+4}{b}_n+b_{n+2}{b}_{n+1}{b}_{n+3}{c_{n+1}}^2{c}_n{c_{n+3}}^2{b}_n-\hfill \\ b_{n+2}{b}_{n+1}{b_{n+3}}^2{c_{n+1}}^2{c}_n{b}_{n+4}{b}_n-b_{n+2}{c_{n+1}}^2{c_n}^3{c_{n+3}}^2{b}_{n+3},\hfill \\ \\ t_n={b_{n+3}}^2{b}_{n+4}{c_{n+1}}^3{c_n}^2{c}_{n-1}-b_{n+3}{c_{n+3}}^2{c_{n+1}}^3{c_n}^2{c}_{n-1}.\hfill \end{array}$$

Note that the hypothesis, $p_n\ne 0$ for all n, is equivalent to $t_n\ne 0$ for all n. Then, for any given pair of linearly independent solutions $A_n,B_n$ of (29), the sequences $A_{n+1}{A}_n{A}_{n-1}$, $B_{n+1}{B}_n{B}_{n-1}$, $A_{n+1}{A}_n{B}_{n-1}+A_{n+1}{B}_n{A}_{n-1}+B_{n+1}{A}_n{A}_{n-1}$, and $B_{n+1}{B}_n{A}_{n-1}$ + $B_{n+1}{A}_n{B}_{n-1}$ + $A_{n+1}{B}_n{B}_{n-1}$ form a linearly independent set of solutions for (37). Given that we know how to test for degree 2 irrationality of limits L via Theorem 5, we can formulate an analogous result for degree 3 irrationality next.

Note: In the sequel, we always assume that the $A_n,B_n$ in question are positive for all n (as they arise from a disconjugate Equation (29)). There is no loss of generality in assuming this, since the proofs involve limiting arguments. In addition, unless otherwise specified, we assume that $L\ne 0$.

Theorem 6. 

Let$c_n,b_n\in \mathbb{R}$and$p_n\ne 0$for all n. Let$A_n,B_n$be two independent rational valued solutions of (29) such that

$$\underset{n\to \infty }{lim}\frac{B_n}{A_n}=L,$$

where L is irrational and L is not algebraic of degree 2. Then, L is algebraic of degree three over$\mathbb{Q}$if and only if (37) has a non-trivial rational valued solution$S_n$such that

$$\frac{S_n}{A_{n+1}{A}_n{A}_{n-1}}\to 0,n\to \infty .$$

(38)

The proof is similar to that of Theorem 4 and so is omitted.

Remark 5. 

A re-examination of the proof of Theorem 4 which serves as a template for all other such proofs to follow shows that the tacit assumptions on L can be waived to some extent. The previous result may then be re-formulated as follows.

Let $S_n$, a solution of (37), have the basis representation

$$\begin{array}{ccc}\hfill & & S_n=a_3{A}_{n+1}{A}_n{A}_{n-1}+a_0{B}_{n+1}{B}_n{B}_{n-1}+\hfill \\ & & a_2(A_{n+1}{A}_n{B}_{n-1}+A_{n+1}{B}_n{A}_{n-1}+B_{n+1}{A}_n{A}_{n-1})+\hfill \\ & & a_1(B_{n+1}{B}_n{A}_{n-1}+B_{n+1}{A}_n{B}_{n-1}+A_{n+1}{B}_n{B}_{n-1}),\hfill \end{array}$$

(39)

where $a_i\in \mathbb{R}$ and the subscript i in $a_i$ for the basis coordinates is determined by counting the number of A values in the basis vector immediately following it.

Theorem 7. 

Let$c_n,b_n\in \mathbb{R}$and$p_n\ne 0$for all n. Let$A_n,B_n$be two independent rational valued solutions of (29) such that

$$\underset{n\to \infty }{lim}\frac{B_n}{A_n}=L.$$

Then, L is algebraic of degree at most 3 if and only if there exists a non-trivial rational valued solution$S_n$of (37) satisfying (38).

Proof of Theorem 7. 

Idea: Using (39), we see that since $S_n$ is rational, then so are the $a_i$, $0\le i\le 3$, not all of which are zero. Next, as $n\to \infty$,

$$\frac{S_n}{A_{n+1}{A}_n{A}_{n-1}}\to a_0{L}^3+3a_1{L}^2+3a_{2}L+a_3,$$

and so L is algebraic of degree no greater than 3. Conversely, let L be algebraic of degree no greater than 3 and let $p\left(x\right)=a_0{x}^3+3a_1{x}^2+3a_{2}x+a_3$ be its defining polynomial where not all $a_i$ are zero. Then, choosing the solution $S_n$ of (37) in the form (39) with the same quantities $a_i$ that appear as the coefficients of p, we see that since $p\left(L\right)=0$, (38) is satisfied. □

Remark 6. 

In order to improve on Theorem 6, we need to add more to the solution$S_n$appearing therein. For example, it is easy to see that under the same basic conditions on the$A_n,B_n$, if there exists a non-trivial rational valued solution$S_n$of (37) with$a_0\ne 0$satisfying (38), then L is algebraic of degree no greater than 3. On the other hand, if L is algebraic of degree 3, then there exists a non-trivial rational valued solution$S_n$of (37) with $a_0\ne 0$satisfying (38).

Theorems 5 and 6 give us an idea on how to proceed next. In essence, we now have some way of determining whether or not the limit L is algebraic of degree 3 based on the fact that it is not algebraic of lower degree. The general result is similar, but first, we describe the construction of the required linear higher-order recurrence relations. In order to test whether the limit L in Theorem 6 is algebraic of degree m, $m\ge 2$, we will require a linear recurrence relation containing $(m+2)$-terms or equivalently an $(m+1)$-th order linear difference equation (an equation involving “finite differences” in the traditional sense). This new equation is found from a prior knowledge of the kernel of the associated operator.

As usual, we let $A_n,B_n$ be two linearly independent solutions of (29). We seek a homogeneous linear $(m+2)$-term recurrence relation whose basis (consisting of $(m+1)$ terms) is described as follows: Two basic elements are given by

$$A_{n+m-2}{A}_{n+m-1}\dots A_n{A}_{n-1}$$

along with a corresponding term with all these A values replaced by B values. For each given k, $0<k<m$, consider

$$\sum A_{n+m-2}{A}_{n+m-1}\dots B_{n+m-i}\dots B_{n+m-j}\dots A_n{A}_{n-1}$$

where the sum contains $\left(\genfrac{}{}{0pt}{}{m}{k}\right)$ distinct terms. Each summand is found by enumerating all the ways of choosing k-terms from the full product of A values and then replacing each such A with a B (keeping the subscripts the same).

As an example, consider the case $m=4$ and $k=2$. There is then such a sum of $6=\left(\genfrac{}{}{0pt}{}{4}{2}\right)$ terms, the totality of which looks like

$$\begin{array}{ccc}& & x_3^{(n-1)}=A_{n+2}{A}_{n+1}{B}_n{B}_{n-1}+A_{n+2}{B}_{n+1}{B}_n{A}_{n-1}+B_{n+2}{B}_{n+1}{A}_n{A}_{n-1}+\hfill \\ & & B_{n+2}{A}_{n+1}{B}_n{A}_{n-1}+A_{n+2}{B}_{n+1}{A}_n{B}_{n-1}+B_{n+2}{A}_{n+1}{A}_n{B}_{n-1}.\hfill \end{array}$$

As k varies from 0 to m, the collection of all such “sums of products” gives us a collection of $(m+1)$ terms of the form $x_1^{(n-1)},x_2^{(n-1)},\dots ,x_{m+1}^{(n-1)}$. That this specific set of elements is a linearly independent set may depend on the nature of the interaction of the $a_n,b_n$ in (29) as we saw above (e.g., $p_n\ne 0$ in (37)). At any rate, since every solution $z_{n-1}$ of this new recurrence relation must be a linear combination of $x_i^{(n-1)}$, we see that the compatibility relation is found by setting the determinant of this matrix,

$$\left[\begin{array}{ccccc}{z}_{n-1}& x_1^{(n-1)}& x_2^{(n-1)}& \dots & x_{m+1}^{(n-1)}\\ z_n& x_1^{\left(n\right)}& x_2^{\left(n\right)}& \dots & x_{m+1}^{\left(n\right)}\\ \dots & \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots & \dots \\ z_{n+m}& x_1^{(n+m)}& x_2^{(n+m)}& \dots & x_{m+1}^{(n+m)}\\ \end{array}\right],$$

equal to zero, for every n. This and the repeated use of the recurrence relation (29) gives the required $(m+2)$-term recurrence relation of which (37) and (30) are but special cases.

Note: In the sequel we always assume that the set consisting of the “sums of products” described above is a linearly independent set of $(m+1)$ elements. This is equivalent to various conditions to be imposed upon the coefficient of the leading and trailing terms of the ensuing $(m+2)$-term recurrence relation whose construction is presented above.

Theorem 8. 

Let$c_n,b_n\in \mathbb{R}$, $c_n\ne 0$and$b_n\ne 0$in addition to other conditions enunciated in the note above. Let$m\ge 3$. Consider two independent rational valued solutions$A_n,B_n$of (29) such that

$$\underset{n\to \infty }{lim}\frac{B_n}{A_n}=L,$$

where L is not algebraic of degree less than or equal to$m-1$. Then, L is algebraic of degree m over$\mathbb{Q}$if and only if the$(m+2)$-term linear recurrence relation described above has a non-trivial rational valued solution$S_n$such that

$$\frac{S_n}{A_{n+m-2}\dots A_{n+1}{A}_n{A}_{n-1}}\to 0,n\to \infty .$$

(40)

An analog of Theorem 7 can also be formulated, which is perhaps easier to use in practice.

Theorem 9. 

Let$c_n,b_n\in \mathbb{R}$, $c_n\ne 0$and$b_n\ne 0$in addition to other conditions enunciated in the note above. Let$m\ge 3$. Let$A_n,B_n$be two independent rational valued solutions of (29) such that

$$\underset{n\to \infty }{lim}\frac{B_n}{A_n}=L.$$

Then, L is algebraic of degree at most m over$\mathbb{Q}$if and only if the$(m+2)$-term linear recurrence relation described above has a non-trivial rational valued solution$S_n$such that

$$\frac{S_n}{A_{n+m-2}\dots A_{n+1}{A}_n{A}_{n-1}}\to 0,n\to \infty .$$

(41)

Remark 7. 

Since the condition in Theorem 9 puts a bound on the degree m of algebraic irrationality over$\mathbb{Q}$, it also gives an equivalent criterion for the transcendence of numbers L whose limits are found by using quotients of solutions of three-term recurrence relations. In particular, associated to the special number$\zeta \left(3\right)$is an infinite sequence of specific linear recurrence relations of every order, as constructed above, involving sums of products of both sets of Apéry numbers$A_n,B_n$. The transcendence of$\zeta \left(3\right)$is then equivalent to the statement that none of the infinite number of (disconjugate) recurrence relations constructed has a non-trivial rational valued principal solution of the type described.

Example 2. 

In this final example, we interpret Apéry’s construction [1], for the irrationality of$\zeta \left(2\right)$in the context of the non-existence of rational valued solutions of recurrence relations with predetermined asymptotics. Recall that Apéry’s three-term recurrence relation for the proof of the irrationality of$\zeta \left(2\right)$is given by [1]

$${(n+1)}^2{y}_{n+1}-n^2{y}_{n-1}=(11n^2+11n+3)y_n,n\ge 1.$$

In order to apply Theorem 2, we need to express this equation in self-adjoint form; that is, we simply multiply both sides by${(-1)}^n$resulting in the equivalent Equation (29) with$c_n={(-1)}^n{(n+1)}^2$, $b_n={(-1)}^n(11n^2+11n+3)$. The Apéry solutions of this equation (e.g., [26]) are given by

$$A_n^{\prime }=\sum _{k=0}^n{\left(\genfrac{}{}{0pt}{}{n}{k}\right)}^2\left(\genfrac{}{}{0pt}{}{n+k}{k}\right),$$

(42)

and

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

(43)

We know, from Apéry’s paper, that these solutions are such that

$$B_n^{\prime }/A_n^{\prime }\to \zeta \left(2\right)$$

as$n\to \infty$, and we already know that$\zeta \left(2\right)$is irrational. It follows from the above considerations that the four-term recurrence relation

$$\begin{array}{c}\hfill p_n{z}_{n+2}+q_n{z}_{n+1}+r_n{z}_n+s_n{z}_{n-1}=0,\end{array}$$

where

$p_n={\left(n+3\right)}^2{\left(n+2\right)}^4\left(11n^2+11n+3\right),$

$q_n=-\left(11n^2+11n+3\right)\left(122n^4+976n^3+2873n^2+3684n+1741\right){\left(n+2\right)}^2,$

$r_n=-\left(11n^2+55n+69\right)\left(122n^4+488n^3+677n^2+378n+76\right){\left(n+1\right)}^2,$

$s_n={\left(n+1\right)}^4{n}^2\left(11n^2+55n+69\right),$and whose solution space is spanned by the elements$A_n^{\prime }{A}_{n-1}^{\prime }$, $B_n^{\prime }{B}_{n-1}^{\prime }$and$A_n^{\prime }{B}_{n-1}^{\prime }+B_n^{\prime }{A}_{n-1}^{\prime }$cannot have a non-trivial rational valued solution$S_n$satisfying

$$\frac{S_n}{A_n^{\prime }{A}_{n-1}^{\prime }}\to 0,n\to \infty .$$

However, we also know that$\zeta \left(2\right)$is actually transcendental (as it is a rational multiple of${\pi }^2$) and so cannot be algebraic of any finite degree. Hence, for each m, none of the$(m+2)$-term recurrence relations that can be constructed as described above has a non-trivial rational valued solution satisfying (38).