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Source. Theorem 3.7, printed pp. 88--89 (statement read on the page images); proof pp. 89--91 with Selberg's Theorem 3.10 on p. 90 (read in the text layer only, which garbles formulas; the structure below follows the prose; (3.9) and (3.11) were checked on the page image of p. 90).

Statement

Suppose the positive integers $a_n$ are monotonic and, for some $\delta>0$, exceed $n^{1/2+\delta}$ once $n$ is large. Then $1,x,y,z$ are linearly independent over $\mathbb Q$, where

$$x=\sum_{n=1}^{\infty}\frac{\varphi(n)}{a_1\cdots a_n},\qquad y=\sum_{n=1}^{\infty}\frac{\sigma(n)}{a_1\cdots a_n},\qquad z=\sum_{n=1}^{\infty}\frac{d_n}{a_1\cdots a_n},$$

with $d_n$ any integers such that $|d_n|<n^{1/2-\delta}$ once $n$ is large and $d_n\ne0$ infinitely often.

Proof structure (pp. 89--91)

Suppose some integers $A,B,C$, not all $0$, make $Ax+By+Cz$ an integer; it is the sum $S=\sum b_n/(a_1\cdots a_n)$ with $b_n=A\varphi(n)+B\sigma(n)+Cd_n$. Since Theorem 2.1 alone shows $z$ irrational, $A$ and $B$ are not both $0$. Case $A+B=D>0$ (after a sign change): Theorem 2.1 gives integers $c_n$ with $b_n=c_na_n-c_{n+1}$, $|c_n|<n^{(1-\delta)/2}$; restricting to prime indices $n=p_m$, where $b_{p_m}=Dp_m+d'_m$ with $d'_m=Cd_{p_m}-A+B$ small, the ratio $p_{m+1}/p_m$ is compared with $c'_{m+1}/c'_m$ (formula (3.8)), so that $c'_{m+1}>c'_m$ forces a gap $p_{m+1}-p_m$ larger than $\sqrt{p_m}$ by a power of $p_m$ ((3.9) prints $p_{m+1}>p_m+\tfrac12p_m^{1/2+\delta}$; the bound $|c_n|<n^{(1-\delta)/2}$ of p. 89 yields only $\tfrac12p_m^{1/2+\delta/2}$). Selberg's theorem (Theorem 3.10, [3, Theorem 4]: for $\Phi(x)=x^{1/2+\delta}$, almost all intervals $(x,x+\Phi(x))$ contain about $\Phi(x)/\log x$ primes) shows such gaps are rare, which yields (3.11) $|c_n|<n^\varepsilon$ for all large $n$ (by the monotonicity of $a_n$) and then that $c'_m$ is eventually constant, $=c$. Comparing consecutive relations at $p$ and $p+1$ shows that the limit points of $(A\varphi(p+1)+B\sigma(p+1))/(D(p+1))$ over primes $p$ would be rationals with denominator $c$ (3.13); Dirichlet's theorem makes $\sigma(p+1)/(p+1)$ dense in $(1,\infty)$ (and $\varphi(p+1)/(p+1)$ dense in $(0,1)$ when $B=0$), a contradiction. Case $A+B=0$: the same argument along the indices $2p$.

Relation to problem 252

With $a_n=n$ (monotone; $n>n^{1/2+\delta}$ for $\delta<1/2$): the numbers $1$, $\sum\varphi(n)/n!$, $\sum\sigma(n)/n!$ and $\sum d_n/n!$ are rationally independent; in particular $\sum\sigma(n)/n!$ is irrational, which is problem 252 for $k=1$. The irrationality alone is already Theorem 1.1 of the card, that is the authors' 1971 result with $a_n=n\ge n^{11/12}$; the 1974 contribution is the linear independence. Hančl–Tijdeman 2005 (p. 2) and 2010 (p. 2) cite this theorem for the independence of $1$, $\sum\sigma(n)/n!$, $\sum\varphi(n)/n!$ and $\sum b_n/n!$ with $|b_n|<n^{1/2-\varepsilon}$, and Hančl–Tijdeman 2010 credits the cases $k=0,1$ of $\sum\sigma_k(n)/n!$ to this paper. Nothing here concerns $\sigma_k$ for $k\ge2$.

Bears on. #252 (the case $k=1$, as a consequence of the rational independence; no bearing on $k\ge2$).