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Source. Theorem 3.1 and its proof, preprint pp. 7--8; Corollary 3.1, p. 8; Proposition 3.1, p. 5, with its proof on pp. 5--6. Read on the rendered pages. The edition read is identified on the source card.

Statement

Let 0<δ<10<\delta<1, let (R(n))n≥1(R(n))_{n\ge1} be a sequence of positive integers and (an)n≥1(a_n)_{n\ge1} a sequence of integers. Suppose that for infinitely many N∈NN\in\mathbb{N} the terms

aN−2R(N), aN−2R(N)+1, …, a⌈N+5R(N)/(1−δ)⌉a_{N-2R(N)},\ a_{N-2R(N)+1},\ \ldots,\ a_{\lceil N+5R(N)/(1-\delta)\rceil}

form a geometric progression, and that

aN+n=o(NR(N)+δn)(n=0,1,…)andN−2R(N)→∞(N→∞).a_{N+n}=o\bigl(N^{R(N)+\delta n}\bigr)\quad(n=0,1,\ldots) \qquad\text{and}\qquad N-2R(N)\to\infty\quad(N\to\infty).

Then S=∑n=1∞an/n!∉QS=\sum_{n=1}^{\infty}a_n/n!\notin\mathbb{Q}.

Here ⌈x⌉\lceil x\rceil is the least integer at least xx (p. 7).

Corollary 3.1 (p. 8) is the case δ=1/6\delta=1/6: with (R(n))(R(n)) and (an)(a_n) as above, if for infinitely many NN the terms aN−2R,…,aN+6Ra_{N-2R},\ldots,a_{N+6R} form a geometric progression and aN+n=o(NR(N)+n/6)a_{N+n}=o(N^{R(N)+n/6}) for n=0,1,…n=0,1,\ldots, with N−2R(N)→∞N-2R(N)\to\infty, then S∉QS\notin\mathbb{Q}.

Scope of the hypothesis (an observation of this page, not of the paper). The proof writes the ratio as aN+1/aN=c/da_{N+1}/a_N=c/d with cc, dd coprime positive integers, so it covers only runs of nonzero terms with positive ratio. A run of zeros meets the printed wording, and a sequence that is zero from some index on gives a rational SS, so zero runs must be excluded. Remark 3.1 (p. 7) says the argument also works for a negative ratio c/dc/d under the extra inequality −cd⋅K+1a(N+K+1)+b<12-\frac cd\cdot\frac{K+1}{a(N+K+1)+b}<\frac12, in the notation of Proposition 3.1 below.

Proof pointer

The theorem is derived from Proposition 3.1 (p. 5). That proposition takes integers a>0a>0, b≥0b\ge0 and integers ana_n such that for infinitely many NN the run aN−K,…,aN+K+Ha_{N-K},\ldots,a_{N+K+H} is a nonzero geometric progression with ratio c/dc/d, c=c(N)c=c(N) and d=d(N)d=d(N) coprime positive integers, K=K(N)K=K(N), H=H(N)H=H(N), N−K→∞N-K\to\infty; under the two growth conditions (5) and (6) it concludes that ∑nan/(a+b)a,n∉Q\sum_n a_n/(a+b)_{a,n}\notin\mathbb{Q}, where (x)a,n=x(x+a)⋯(x+(n−1)a)(x)_{a,n}=x(x+a)\cdots(x+(n-1)a). The proof forms an integer combination DND_N of tails of the series with KK-th difference weights, evaluates it with the summation identity of Lemma 2.3 (p. 4), and plays a lower bound for ∣DN∣|D_N| against its divisibility by K!/AKK!/A_K. Theorem 3.1 takes a=1a=1, b=0b=0, K=2RK=2R and H=⌈3+2δ1−δR⌉H=\lceil\frac{3+2\delta}{1-\delta}R\rceil with R=R(N)R=R(N), and checks (5) and (6) from the growth bound (pp. 7--8).

Dependencies

Proposition 3.1, Lemma 2.1 and Lemma 2.3 of the same paper.

Bears on

No catalog problem directly. It is the tool behind Corollary 3.2.