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Vandehey 2012 incomplete argument erdos irrationality lambert series

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theorem_1_1: Vandehey's statement that for an integer b > 1 and a finite set A of non-negative integers, the sum of d(n) a_n/b^n is irrational for every sequence with values in A that does not end in repeated zeros; the paper says Erdős's method gives it with a virtually identical proof.

theorem_1_2: Vandehey's theorem that for an integer b > 1 and a finite set A of integers not containing 0, the sum of d(n) a_n/b^n is irrational for every sequence with values in A; the choice a_n = (-1)^n gives irrationality of the divisor Lambert series at 1/c for every integer c < -1.


Joseph Vandehey, On an incomplete argument of Erdős on the irrationality of Lambert series. arXiv preprint (2012). arXiv:1206.0340.

The paper shows that the Lambert series f(x)=∑d(n)xnf(x)=\sum d(n)x^n is irrational at x=1/bx=1/b for every negative integer b<−1b<-1 (abstract, p. 1), by an elementary argument that repairs a gap in Erdős's original proof: Erdős proved irrationality of f(1/b)f(1/b) for integers b>1b>1, and for b<−1b<-1 his method gives arbitrarily long strings of 00's in the base ∣b∣|b| expansion, but he claimed without proof that the expansion does not end in 00's (p. 1). Theorem 1.1 (p. 2), which Vandehey says Erdős's method gives with a virtually identical proof and does not prove in the paper, states that for an integer b>1b>1 and a finite set A\mathcal A of non-negative integers, ∑d(n)an/bn\sum d(n)a_n/b^n is irrational for every sequence (an)(a_n) with values in A\mathcal A that does not end in repeated 00's. Theorem 1.2 (p. 2), proved in Section 2 (pp. 2--5), states the same for every sequence with values in a finite set A\mathcal A of integers not containing 00; taking an=(−1)na_n=(-1)^n gives the case b<−1b<-1 (the sentence on p. 2 prints "b<1b<1" [sic]). The new ingredient is finding arbitrarily long strings of zeros in the base-bb expansion that are preceded by a non-zero digit, arbitrarily far out. The proof uses a lower bound for primes in arithmetic progressions that the paper says is mentioned by Alford, Granville and Pomerance (Proposition 2.1, pp. 2--3), and a tail estimate of Erdős given without proof (Lemma 2.2, p. 4). The proof extends Erdős's method; the paper says the later proofs of the case b<−1b<-1 use entirely different techniques (p. 1). For problem 1049, which asks about rational t>1t>1, Theorem 1.2 with an=1a_n=1 recovers Erdős's case of integer t>1t>1; the negative-base result completed here lies outside the problem's range t>1t>1.

Source: https://arxiv.org/abs/1206.0340. The arXiv record names arXiv's non-exclusive distribution license (arXiv:1206.0340), every other right reserved.

The copy read for this card is arXiv:1206.0340v1 (2 June 2012).

Read status: claims checked for Theorems 1.1 and 1.2, read clause by clause on the page images of the print; the proof of Theorem 1.2 in Section 2 read for structure, not verified. Proposition 2.1 and Lemma 2.2 are cited in the paper without proof and were not checked. Nothing here is independently reviewed.

Bears on. #1049: Theorem 1.2 (p. 2) with an=1a_n=1, like Theorem 1.1 with A={1}\mathcal A=\{1\}, gives irrationality for every integer t>1t>1, the case the problem credits to Erdős; the paper's new case is a negative integer base, outside the problem's range, and it says nothing about non-integer rational tt.

Results.

  • Theorem 1.1 (p. 2): for an integer b>1b>1 and a finite set A\mathcal A of non-negative integers, ∑n≥1d(n)an/bn\sum_{n\ge1}d(n)a_n/b^n is irrational for every sequence (an)(a_n) with values in A\mathcal A that does not end in repeated 00's.
  • Theorem 1.2 (p. 2): for an integer b>1b>1 and a finite set A\mathcal A of integers not containing 00, ∑n≥1d(n)an/bn\sum_{n\ge1}d(n)a_n/b^n is irrational for every sequence (an)(a_n) with values in A\mathcal A; with an=(−1)na_n=(-1)^n the sum is f(−1/b)f(-1/b), so f(1/c)f(1/c) is irrational for every integer c<−1c<-1.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.