Convergence properties of the classical and generalized Rogers-Ramanujan continued fraction

The aim of this paper is to study the convergence and divergence of the Rogers-Ramanujan and the generalized Rogers-Ramanujan continued fractions on the unit circle. We provide an example of an uncountable set of measure zero on which the Rogers-Ramanujan continued fraction $R(x)$ diverges and which enlarges a set previously found by Bowman and Mc Laughlin. We further study the generalized Rogers-Ramanujan continued fractions $R_a(x)$ for roots of unity $a$ and give explicit convergence and divergence conditions. As such, we extend some work of Huang towards a question originally investigated by Ramanujan and some work of Schur on the convergence of $R(x)$ at roots of unity. In the end, we state several conjectures and possible directions for generalizing Schur's result to all Rogers-Ramanujan continued fractions $R_a(x)$.


Introduction
An infinite continued fraction is an expression of the form where a i and b i can be real or complex numbers, or functions of one or several variables, as will be the case throughout this paper. For space considerations the following notation is used: We further adopt the notations: As usual, we let P n and Q n be the unreduced numerator and denominator of the n-th convergent (or approximant) of the continued fraction, that is, It is known (see, e.g., [5, pp. 9]) that, for n ≥ 2, P n and Q n satisfy the following recursions: (2) P n = b n P n−1 + a n P n−2 , Q n = b n Q n−1 + a n Q n−2 and also that, for n ≥ 1, x 1 + The Rogers-Ramanujan continued fraction is known for its connections with the celebrated Rogers-Ramanujan identities: , where (a; q) 0 := 1, (a; q) n := (1 − aq j ), |q| < 1.
For our purposes we shall use the former, namely We introduce For a survey on the Rogers-Ramanujan continued fraction and Rogers-Ramanujan identities, we refer the reader to [2,Section 1] and [3, pp. 3325-3327].
We are interested in studying the convergence of the Rogers-Ramanujan continued fraction as well as its generalization. Let us recall that the continued fraction K ∞ n=1 an bn is said to converge if Pn Qn converge as n → ∞. It is clear that the convergence behavior of R(x) (and R a (x)) is the same with that of K(x) (and K a (x)). It is immediate by the following classical theorem that R(x) (hence also K(x)) converges to a value in C for all |x| < 1.
Theorem 1 (Worpitzky, [5]). Let the continued fraction K ∞ n=1 a n /1 be such that |a n | ≤ 1/4 for n ≥ 1. Then K ∞ n=1 a n /1 converges. All approximants of the continued fraction lie in the disk |w| < 1/2, and the value of the continued fraction is in the disk w ≤ 1/2. It is natural, however, to consider the expressions (4) and (5) also for |x| ≥ 1 and ask questions about convergence. Andrews et al. [1] established what was Ramanujan's original claim, showing that for |x| < 1 we have Thus, the interesting case to study convergence properties is when |x| = 1, which is a much more subtle question. Some work has been carried out in this regard. In an important 1917 paper of Schur [9], he showed that if x is a primitive m-th root of unity and m ≡ 0 (mod 5), then K(x) diverges, and if m ≡ 0 (mod 5), then K(x) converges with where λ = m 5 is the Legendre symbol and σ the least positive residue of m modulo 5. It has been an open question since Schur's paper whether the Rogers-Ramanujan continued fraction converges or diverges at points on the unit circle which are not roots of unity. This convergence question was utilized by Lubinsky to provide an important counterexample to a conjecture of Baker-Gammel-Wills in [6], and Bowman and Mc Laughlin performed a comprehensive study of convergence at roots of unity in [3]. In particular, they showed the following.
Theorem 2 (adapted from [3]). Define where [0; e 1 (t), e 2 (t), . . .] denotes the continued fraction expansion of t and c i (t)/d i (t) its i-th convergent. Then S is an uncountable set of measure zero, and if t ∈ S and y = exp(2πit) then K(y) diverges.
In this paper we improve this result by proving Then S R is an uncountable set of measure zero, and if x ∈ S R then K(x) diverges.
Theorem 3 improves Theorem 2, in the sense that the set S R extends the set S locally around −1 for any R ∈ (0, √ 5−1). The smaller one chooses R, the larger the set is around −1 on the unit circle.
Example. We now give an example of y ∈ S R \ S. For this, choose e 1 = 1, e 2 = 1, e 3 = 2, e n+1 = λ dn/2 R for n ≥ 3; The computation will be detailed in Section 2.
Remark. Note that the condition t ∈ R \ Q is essential in both Theorem 2 and 3, because we need an infinite continued fraction expansion for t.
Despite the effort which has been expended on studying the convergence of the Rogers-Ramanujan continued fraction, very little is currently known about other continued fractions. As a generalization of what is known for the Rogers-Ramanujan continued fraction, a natural first step is to determine at which roots of unity does the generalized Rogers-Ramanujan continued fraction converge. We provide an extension of Schur's result, which in particular explains the 5-divisibility condition in his theorem as part of a larger framework.
The limit we compute in the following theorem coincides with what Ramanujan originally claimed in [7, pp. 57]. While stating the limit, he leaves open the question of when does it actually exist. Huang partially answers this question in [4] by using mostly analytic arguments. Here we come up with a purely algebraic approach which leads to different conditions; however, whenever Huang's and our conditions intersect, the results are consistent. If m ∈ N \ {2}, K a converges at x = ζ m to some limit in Q a, 1 4 + a m , ζ m * ⊂ C * if and only if 1 4 + a m / ∈ R ≤0 In this case, the limit is given by: where P −1 (a, ζ 1 ) := 1 and P −2 (a, ζ 1 ) := 0. If m = 2, that is, x = −1, then K a diverges.
Theorem 5. Let ζ m be a primitive m-th root of unity and a ∈ C. Assume 1 4 + a m ∈ R <0 . Then the generalized Rogers-Ramanujan continued fraction is divergent at x = ζ m .
Note now that Schur's result regarding the convergence of K(x) in the case m ≡ 0 (mod 5) follows as a consequence of Theorem 4 upon setting a = 1. Briefly, the paper is organized as follows. In Section 2 we prove Theorem 3. The proof requires several preliminary steps which are similar in spirit with the ideas in [3]. In Section 3 we give the proofs of Theorems 4 and 5. We conclude by giving a complete conjectural description of the convergence and divergence at roots of unity.

Divergence of the Rogers-Ramanujan continued fraction
In this section we study the divergence of the Rogers-Ramanujan continued fraction for |x| = 1 and prove Theorem 3. The following lemma will be a key step.
with unreduced numerator P N and denominator Q N converges, then Proof. This is essentially proved in [3, pp. 3331] but here we give a self-contained proof for the reader's convenience. By (3) we have yielding the desired claim.
Next, we recall the algorithm used to compute the continued fraction expansion of an irrational number in (0, 1). Let t ∈ (0, 1) \ Q. We define the recursive sequences: If c n (t) and d n (t) denote the n-th numerator and denominator, then by (3) it is immediate that gcd(c n (t), d n (t)) = 1.

Lemma 2. With the previous notations, we have
Proof. By the definitions and the recursion relation, we have Now, for K(x) we have the recursion relations (Note that defining the initial terms P −1 , P 0 and Q −1 , Q 0 does not affect the recursion.) We adapt an observation from Rogers in [8, Lemma I] and recombine any three consecutive equations to obtain new recursion formulae: with initial terms As we decoupled the even and odd indexed convergents, it is now convenient to introduce a new notation: P n,k := P 2n+k , Q n,k := Q 2n+k for n ∈ N and k ∈ {0, 1}.
The following will provide a crucial bound for our computation.
there exists a constant depending on R, say C(R) > 0, such that , therefore the general term will be equal to α n = a · λ n R + b · λ ′n R , for some constants a, b.
Lemma 3 applies to our problem in the following sense. Let M R = {x ∈ C : |x| = 1, |x + 1| < R} be as in Section 1.
Proof. Apply Lemma 3 to the two sequences {Q n,0 (x)} n≥0 and {Q n,1 (x)} n≥0 separately. We only consider {Q n,0 (x)}. For the sequence {α n } n≥0 in Lemma 3 choose the initial values: Then, by induction, one easily sees that, for n ≥ 2, Lemma 5. For any R > 0 there exist A(R) > 0 and µ ≥ 0, such that: Proof. Let x, y ∈ M R . To satisfy the claim for n ∈ {−1, 0, 1, 2}, we only need to find the conditions Thus, to verify the base case, we only need to bound A(R) and µ from below. It is again necessary to prove the assertion for even and odd n separately. Without loss of generality we deal only with the even case. Letting n ≥ 2, we compute: where the last inequality follows from Lemma 4. In order to prove the assertion, we assume the induction hypothesis to hold for n − 1 and n − 2 and seek the sufficient conditions for µ and A(R), for all n. We need the following equation to hold: Luckily, the terms of order (n+µ) 2 cancel on both sides and by equivalent elementary manipulations it reduces to: But since we still have the freedom to choose A(R) and µ very large, this condition can be satisfied for all n. Since the same calculation is possible for the odd sequence, the statement holds.
We finally prove the main divergence result.
Then S R is an uncountable set of measure zero, and if x ∈ S R then K(x) diverges.
Proof. Let x = e 2πit ∈ S R . Again, c n (t), d n (t) denote the unreduced numerators and denominators of the continued fraction expansion of t ∈ (0, 1). Define x n := e 2πicn(t)/dn(t) . Since gcd(c n (t), d n (t)) = 1, x n is a primitive d n (t)-th root of unity. Let n ∈ N. As a direct consequence of the explicit values computed by Schur in [9, pp. 134] we find Now, for x n ∈ M R , in turn by Lemma 5, the fact that chord length is shorther than arc length and Lemma 2, we have and a similar result for d n (t) − 2. As such, since t ∈ T R , for infinitely many n, the right-hand side is bounded by a finite constant, say B > 0. Then, for these n, using the triangle inequality and inequalities (18) and (19), we see that and the proof concludes by Lemma 1. It is clear that S R is an uncountable set and the fact that it has measure zero follows from [3, Lemma 3].
Example. We construct an example with x ∈ S R \ S. As for s, t ∈ R, e 2πis − e 2πit ≤ 2π |s − t| , we need x = e 2πit with t close to 1 2 . As Hence, x = e 2πit ∈ S R . Because λ 1 2

Convergence of the Generalized Rogers-Ramanujan Continued Fraction
In this section we study the convergence of the generalized Rogers-Ramanujan fraction at roots of unity. More precisely, we study the convergence of K a (x) = 1 1 + aR a (x).

3.1.
Preparations. Before we start, we make a small adjustment of the approximants P n , Q n of K a , as compared to the definition given in (1), which will prove to be more convenient for our computations; namely, we let P n (a, x) Q n (a, x) = 1 1 + n K i=0 ax n 1 .
We will need a special class of determinants closely related to the approximants of the numerator and denominator.
Definition 1 ([9, pp. 120]). Let x 1 , . . . , x m be formal variables or complex numbers. Then define: We want to summarize some of the properties of these determinants. The following is due to Perron, but also cited in [9]. For the sake of completeness, and as the original source is difficult to find, we include the proof here. Proof. The first formula is obtained immediately by using the Laplace expansion on the last row or column. For the second formula we take a closer look into the combinatorics. The matrix in (20) has tridiagonal form, i.e., only the entries on the main diagonal and the first diagonals above and below are non-zero. Then, in the definition of the determinant, σ ∈ S n+1 contributes if and only if for all i ∈ {1, . . . , n + 1} we have |σ(i) − i| ≤ 1. This implies that σ is a composition of disjoint transpositions of neighboring indices. We now distinguish σ by what it does with the index m. Let A = (a i,j ) denote the matrix from (20): With the notations above we get (23) P n (a, x) = D(ax, . . . , ax n ) and Q n (a, x) = D(a, ax, . . . , ax n ) for the n-th approximants of the numerator and denominator. This is because the sequences on both sides of the equations satisfy the same recurrence relation and have the same initial values. These identities then imply a new recurrence relation.
Lemma 7. Let x be an m-th root of unity, m ≥ 3. Then for q ∈ N and r ∈ {0, . . . , m − 1}, we have Proof. We use the representation (23) of the approximants and use (22) to find that P (q+1)m+r (a, x) The same procedure works for Q (q+1)m+r (a, x).
In what follows, m k x denotes the generalized binomial coefficient, that is, Proof. One easily checks that the formulae hold for m ∈ {0, 1}. Since both sequences satisfy the recursion relation the induction step is completed by using the well-known identity (see, for e.g., [9]): It is essential to understand the eigenvalues of the matrix in (24), as they will determine the limit behavior of the approximants P n and Q n . The eigenvalues are determined by the characteristic polynomial. As we have a 2 × 2 matrix, we only need to know the trace and determinant. For x any primitve m-th root of unity, the determinant is easily calculated from (3) to be: Determinig the trace is a rather tricky combinatorial exercise.
Then T m (a, x) has the form

Proof. The proof follows from Lemma 8 and (27).
Remark. The fact that T m (a, x) = 1 for x an m-th root of unity can be alternatively proven by invoking Theorem 2.2 in [4]. Huang [4] considers the following version of the generalized Rogers-Ramanujan continued fraction Let p n , q n be the unreduced n-th numerator and denominator of R(a), for n ≥ 1. Note that If we now let P n , Q n be as in (1), we see that with our notation, for n ≥ 2, we have P n = P n−2 and Q n = Q n−2 . By (30) it easily follows that P n = q n−1 and Q n = p n−1 + q n−1 .
Next, let A n (l) and B n (l) be the subsets of A n and B n which contain all the l-dimensional vectors and C n (l) := A n−1 (l) ∪ B n (l). By the proof of Theorem 2.2 in [4], we showed that follows.
However, this should not come as a surprise, as the sets A n (l) and B n (l) parametrize precisely the permutations that contribute to the determinant formula (23), namely, the n i is the lower index of any transposition of neighboring indices, as explained in the proof of Lemma 6.

3.2.
Proof of the main convergence results. By using these preparatory steps, we prove the following. If m ∈ N \ {2}, K a converges at x = ζ m to some limit in Q a, 1 4 + a m , ζ m * ⊂ C * if and only if 1 4 + a m / ∈ R ≤0 In this case, the limit is given by: where P −1 (a, ζ 1 ) := 1 and P −2 (a, ζ 1 ) := 0. If m = 2, that is, x = −1, then K a diverges.
Proof. Let x = ζ m be a primitive m-th root of unity. If m ≥ 3, the recursion from Lemma 7 holds. In this case we want to analyze the asymptotic behavior of {P qm+r } q and {Q qm+r } q using the matrix Using the previous results, we compute the characteristic polynomial to be: By the assumption on a m , the eigenvalues are different and thus the matrix can be diagonalized. We calculate the eigenvalues and eigenvectors to be: .
Using Lemma 8, we see that the matrix has entries in the field Q(a, ζ m ). By assumption, 1 4 + a m is not in this field. Hence the main diagonal of A m −λ ± I 2 cannot vanish. As the rows and columns of this matrix are multiples of each other, none of the entries can vanish, making the eigenvectors above well-defined. We note, that the eigenvectors are forced to be in Q a, ζ m , 1 4 + a m 2 \ Q(a, ζ m ) 2 . Consider now the eigenvector expansion for r ∈ {0, . . . , m − 1}: =⇒ P qm+r (a, ζ m ) Q qm+r (a, ζ m ) = a r+ λ q As the left-hand side is in Q(a, ζ m ) 2 , none of the coefficients a r± can vanish. We can now finally analyze the convergence behavior. If we use the principal branch to define the complex square root, then |λ + | ≥ |λ − |. Equality is equivalent to 1 4 + a m ∈ R ≤0 . We find for all r ∈ {0, . . . , m − 1} that If |λ + | > |λ − |, the expression is convergent for all r and the limits are equally given by the assertion.
If m = 1, (24) does not hold. Nevertheless, one obtains the same results by analyzing directly the recursion relations R m (a, 1) = R m−1 (a, 1) + aR m−2 (a, 1) for P and Q. This allows to directly calculate the limit for 1 4 + a / ∈ R ≤0 by similar arguments as above. The divergence proof holds equally.
As a partial converse to the previous result, we show the following.
Theorem 5. Let ζ m ∈ C a primitive m-th root of unity and a ∈ C. Assume 1 4 + a m ∈ R <0 . Then the generalized Rogers-Ramanujan continued fraction is divergent at x = ζ m .
Proof. The idea is to establish the inequality (32) independent of the algebraic condition 1 4 + a m / ∈ Q(a, ζ m ) We need to find some r ∈ 0, . . . , m − 1 such that both vectors in the eigenvector expansion do not vanish. The recursion relations for the P n and Q n can be written in vector form: Assume, for some n, that (P n (a, ζ m ), Q n (a, ζ m )) is an eigenvector of the matrix in (24). The vector of index n − 1 cannot be an eigenvector of the same eigenvalue, because the two vectors are linearly independent by the following determinant formula: det P n (a, ζ m ) P n−1 (a, ζ m ) Q n (a, ζ m ) Q n−1 (a, ζ m ) If the index n − 1 vectors were an eigenvector of the other eigenvalue, then, in order to satisfy the recursion relation, the vector of index n − 2 would need to have non-vanishing coefficients in front of both eigenvectors in the eigenvector expansion. Thus, we find an index r ∈ {0, . . . , m − 1} with: Then (32) can be similarly established for this r, proving the divergence. Proof. We have .
Using the well-known fact that the result immediately follows from Theorem 4.
Remark. The containment condition on 1 4 + a m is natural in certain cases. For instance, if a = 1, it identifies exactly the roots of unity where the continued fraction is not convergent, as it coincides with an earlier Theorem by Schur, see [9, pp. 135]. However, there are also cases where it is fulfilled and the fraction is still convergent. For example, let Then there is a unique real algebraic number a ∈ 0, 1 4 with 1 4 + a m = b ∈ Q. But by [4,Theorem 4.4] the fraction is still convergent. The exact places where this condition is necessary remains a mystery.

Open Questions.
The first open question, mentioned above, is to simplify or replace the algebraic condition 1 4 + a m / ∈ Q(a, ζ m ). As mentioned in Remark 3.2, it does not seem entirely natural and could be too strong. Nevertheless, we further examined the case of a being a root of unity, as this is probably the most interesting (or at least, the easiest) one. Although we could not come up with a proof, we can conjecture the exact behavior in this case. Note that only the "⇒" direction is open. The reverse direction "⇐" is explicitly solved (see the proof of Corollary 1 above). The conjecture implies that, for a = ζ k and x = ζ m a root of unity, then 1 4 + a m ∈ Q (a, ζ m ) = Q ζ lcm(k,m) ⇔ a m = 1 and 5 | lcm (k, m) or a m = −1 and 3 | lcm (k, m) .
Here the second condition is not interesting, as this case is already covered by Theorem 5. The first condition can be resolved to 5, k | m. We want to state a specific conjecture for this case that we checked numerically using SAGE for 1 ≤ k ≤ 50 and 3 ≤ m ≤ 100. Then for any r ∈ {s − t : s ∈ R, t ∈ {1, 2}} ∩ N 0 we find ax m−1 P m−3 (a, x) P m−2 (a, x) ax m−1 Q m−3 (a, x) Q m−2 (a, x) P r (a, x) Q r (a, x) = λ P r (a, x) Q r (a, x) for λ = 1 2 ± 1 4 + a m .
Therefore we always find two consecutive indices, such that the vectors are eigenvectors. This implies that K a (x) diverges.
What are the consequences of this conjecture? By the determinant formula (3) two consecutive vectors cannot be linearly dependent. Hence, the vectors must be eigenvectors of different eigenvalues. By the assumption 5, k | m, the eigenvalues are real-valued and have different absolute values. Therefore, the consecutive vectors will produce two different limit points in the sequence {P n /Q n } n , to be specific the ratios of the first and second entry of the linearly independent eigenvectors. Therefore, the continued fraction is divergent. In conclusion, these two conjectures imply the following general picture, which predicts in an explicit manner the set of limit points of K a (x), for any a and x being roots of unity. In case of convergence, the limit is given by (7). If 5, k | m, the two limit points of P n /Q n are exactly P m−2 (a, x) 1 2 ± 1 4 + a m − ax m−1 P m−3 (a, x) ∈ C.
For 3 | m and a m = −1 these limit points can also occur. If {P qm+r /Q qm+r } q does not converge to one of these limit points, it produces exactly these three more limit points and P r (a, x) Q r (a, x) = a r+ v + + a r− v − .