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Source. Lemma 4, printed pp. 215--216, physical PDF pp. 4--5; proof pp. 216--218; Lemma 4′ on p. 218. Read on the page images.

Statement

Let $t>1$ be an integer. Let $a_k$ and $b_k$ ($k=1,2,\ldots$) be sequences of nonnegative integers with infinitely many $a_k>0$. Write $f(n)$ and $g(n)$ for the number of $k$ with $1\le k\le n$ and $a_k>0$, respectively $b_k>0$. Assume:

• (5) there is an $s$ with $a_k<k^s$ and $b_k<k^s$ for all sufficiently large $k$;
• (6) there is an infinite sequence $m_i$ with

$$\sum_{k=1}^{m_i}(a_k+b_k)<c_1m_i,\qquad f(m_i)=o(m_i),\qquad g(m_i)=o\!\left(\frac{m_i}{\log m_i}\right);$$

• (C) for some absolute constant $c_2$: if $i_1<i_2$ are adjacent elements of $\{i:b_i>0\}$ and $x$ satisfies $i_1+c_2x<i_2$, then $a_k>0$ for some $k$ in the open interval $(i_1+x,\,i_1+c_2x)$.

Then, for every choice of signs $\varepsilon_k=\pm1$,

$$\sum_{k=1}^{\infty}\frac{a_k+\varepsilon_kb_k}{t^k}$$

is irrational.

The paper notes (p. 216) that Lemma 1 is the special case with all $b_k=0$ (it prints $m_i=i$; under Lemma 1's $\liminf$ hypothesis $m_i$ must run through indices with $f(m_i)/m_i\to0$).

Structure of the proof (pp. 216--218)

Put $A_k=a_k/t+a_{k+1}/t^2+\cdots$ and $B_k=b_k/t+b_{k+1}/t^2+\cdots$. The lemma follows from (7): for every $\varepsilon>0$ there are indices $j$ with $A_j+B_j<\varepsilon$ and $A_j>B_j$. For if the sum were $u/v$, then (8) $vt^{j-1}\sum_k(a_k+\varepsilon_kb_k)/t^k$ would be an integer, while it also equals $I'+v(A_j+\vartheta B_j)$ with $I'$ an integer and $|\vartheta|\le1$; choosing $\varepsilon<1/v$ and $j$ as in (7) gives $0<v(A_j+\vartheta B_j)<1$, a contradiction.

To prove (7), let $\alpha_i$ count the $k<m_i/2$ with $A_k+B_k\ge\varepsilon$ (9) and $\beta_i$ the $k<m_i/2$ with $A_k>B_k$ (10). It suffices that (11) $\alpha_i=o(m_i)$ and (12) $\beta_i>c_3m_i$.

• (11): split the $k<m_i/2$ satisfying (9) into those within $l$ of an index $j$ with $a_j+b_j>0$, at most $(l+1)(f(m_i)+g(m_i))=o(m_i)$ of them by (6), and the rest, whose values $A_k+B_k$ sum to at most $2c_1m_i/t^l+o(m_i)<\eta m_i$ by (5) and (6) once $l$ is large; the second class therefore has at most $(\eta/\varepsilon)m_i=o(m_i)$ members ((13), (14)).
• (12): if $a_k>0$ and the next index $i>k$ with $b_i>0$ satisfies $i>k+c_4\log k$, then $A_k>B_k$ by (5), since $\sum_{i>k+c_4\log k}i^s/t^{i-k}<1/t$; the same then holds for every $j<k$ with no positive $b$ in $(j,k)$ (15). Let $j<j'$ be consecutive indices with positive $b$. Because $g(m_i)=o(m_i/\log m_i)$, the gaps $j'-j$ exceeding $2c_4\log m_i$ account for $\tfrac12m_i+o(m_i)$ of the range $k<m_i/2$ (16); condition (C) places an index $k_1\le(j+j')/2$ with $a_{k_1}>0$ and $k_1-j>(j'-j)/2c_2$ (17); every $k$ with $j<k\le k_1$ then satisfies $A_k>B_k$ (18), so $\beta_i>(\tfrac12m_i+o(m_i))/2c_2>c_3m_i$ (19).

These steps were read for structure and are recorded as a sketch; the constants $c_1,\ldots,c_4$ are the paper's.

Lemma 4′ (p. 218, stated without proof)

In the setting of Lemma 4 (nonnegative integers $a_k$, $b_k$, infinitely many $a_k>0$, condition (C)), the growth condition (5) can be traded for $\limsup_k(a_k+b_k)^{1/k}<t$ and the last requirement of (6) relaxed to $g(m_i)=o(m_i)$: if some infinite sequence $m_i$ has $\sum_{k\le m_i}(a_k+b_k)<c_1m_i$, $f(m_i)=o(m_i)$ and $g(m_i)=o(m_i)$, then $\sum_k(a_k+\varepsilon_kb_k)/t^k$ is irrational for every choice of signs $\varepsilon_k=\pm1$. The paper says only that "the proof is very similar to that of lemma 4, only the proof of $\beta_i>c_3m_i$ is a bit more troublesome here"; no proof is given there, and none is recorded here.

Role

Used in the proof of Theorem 2 (pp. 218--219).

Bears on. No catalog problem directly; it is the tool behind Theorem 2 and contains Lemma 1 as a special case.