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Source. The remark on printed pp. 377–378, including equation (7), and the introductory comparisons on p. 375 (PDF pp. 1–3). This is a complete rewritten proof of the comparison. Attribution of the earlier condition to Selfridge is as reported by Berger, Felzenbaum and Fraenkel through their reference to Churchhouse; that separate historical source has not been compiled here.

The power-series inequality

For $0\le r_i<1$, put $D=\prod_i(1-r_i)$ and $R=\sum_i r_i$. Then

$$\frac{2-R}{D}\le 2+\sum_i\frac{r_i}{1-r_i}. \tag{1}$$

To prove this, expand the finite product of the absolutely convergent geometric series $\prod_i\sum_{k\ge0}r_i^k$. For a monomial $r_1^{k_1}\cdots r_n^{k_n}$, let $j$ be the number of its positive exponents. Its coefficient on the left of (1) is $2-j$: multiplying by $-r_i$ subtracts one exactly when $k_i>0$. On the right, the constant coefficient is $2$, a monomial supported on one coordinate has coefficient $1$, and every other monomial has coefficient $0$. The coefficients agree for $j=0,1,2$ and satisfy $2-j\le0$ for $j\ge3$. All monomials are nonnegative, so summing proves (1). Absolute convergence justifies the coefficient comparison even when some $r_i=0$.

Let $x_i=r_i/(1-r_i)$ and $F(x)=\prod_i(1+x_i)-\sum_i x_i$. Equation (1) gives

$$F(x)-2 =\frac1D-\sum_i\frac{r_i}{1-r_i}-2 \le\frac{R-1}{D}. \tag{2}$$

Finite and limiting covering conditions

For an odd integer $N=\prod_i p_i^{s_i}$, take

$$r_i=\sum_{j=1}^{s_i}p_i^{-j}.$$

Then $0<r_i<1$ and $x_i=r_i/(1-r_i)$ is exactly the parameter in the product-set theorem. Thus $\sum_{i,j}p_i^{-j}<1$ forces $F(x)<2$ and excludes a cover with distinct cardinalities. Equivalently, such a cover requires $\sum_{i,j}p_i^{-j}\ge1$. Passing to the strictly larger infinite geometric sums shows that an integer covering with these odd prime divisors must satisfy

$$\sum_i\frac1{p_i-1}>1. \tag{3}$$

One can compare the exponent-free conditions directly as well. In (2) take $r_i=1/(p_i-1)$, so $x_i=1/(p_i-2)$ and $D^{-1}=\prod_i(p_i-1)/(p_i-2)$. Then

$$\begin{aligned} &\prod_i\frac{p_i-1}{p_i-2}-\sum_i\frac1{p_i-2}-2\\ &\qquad\le \prod_i\frac{p_i-1}{p_i-2} \left(\sum_i\frac1{p_i-1}-1\right). \tag{4} \end{aligned}$$

The new necessary condition makes the left side positive, so it implies (3). Finally, for nonnegative $u_i=1/(p_i-1)$, $\prod_i(1+u_i)\ge1+\sum_i u_i$. Thus (3) implies the weaker direct-density condition

$$\prod_i\frac{p_i}{p_i-1}>2. \tag{5}$$

For completeness, the latter condition also follows directly from coverage: each distinct modulus is a distinct divisor $m>1$ of $N$, so the density union bound gives

$$1\le\sum_{m\text{ used}}\frac1m \le\sum_{m\mid N,\,m>1}\frac1m =\prod_i\left(\sum_{j=0}^{s_i}p_i^{-j}\right)-1 <\prod_i\frac{p_i}{p_i-1}-1.$$

The strict last inequality uses finite positive exponents. These implications compare necessary conditions only; none is sufficient to construct a cover.

Bears on. The hierarchy of obstructions for Problem 7.