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Source. Liu--Sawhney, On further questions regarding unit fractions, arXiv:2404.07113v1, Lemma 3.1, p. 9; see the source digest.

Statement. Let $N$ be sufficiently large, $N^{0.95}\leq M\leq N$, and $A\subseteq[M,N]\cap\mathbb Z$ with $|A|\geq N^{0.95}$. For each $n\in A$, let

$$(\log N)^{-2}\leq p_n\leq1-(\log N)^{-2}.$$

Let $Q>0$ and choose $x$ so that $x/Q=\sum_{n\in A}p_n/n$. With $e(t)=\exp(2\pi it)$ and the sum over integer frequencies $h$,

$$\frac1Q\sum_{|h|\leq M/2} \operatorname{Re}\!\left( e(-hx/Q)\prod_{n\in A}(1-p_n+p_ne(h/n))\right) \geq\frac3{4Q}.$$

In the later applications $Q$ is a positive integer common denominator and $x$ is an integer. This lemma's proof requires only $Q>0$ and the displayed relation between $x/Q$ and the probabilities.

Proof. Write

$$F(h)=e(-hx/Q)\prod_{n\in A}(1-p_n+p_ne(h/n)), \qquad H=M^{3/5}.$$

First consider $|h|\leq H$. The Taylor estimate in Fact 2.5 gives, uniformly in $n\in A$,

$$1-p_n+p_ne(h/n) =e(p_nh/n) \exp\!\left(-\frac{2\pi^2p_n(1-p_n)h^2}{n^2}\right) \left(1+O\!\left(\frac{|h|^3}{n^3}\right)\right).$$

The conversion of the quadratic polynomial to an exponential introduces an error $O(h^4/n^4)$, which is absorbed by $O(|h|^3/n^3)$ because $|h|/n\leq M^{-2/5}$. The phase factors cancel exactly:

$$e(-hx/Q)\prod_{n\in A}e(p_nh/n)=1.$$

Moreover,

$$\sum_{n\in A}\frac{|h|^3}{n^3} \leq H^3\sum_{n\geq\lceil M\rceil}\frac1{n^3} \ll M^{9/5}M^{-2}=M^{-1/5}.$$

Multiplying the error factors therefore yields

$$F(h)=\bigl(1+O(M^{-1/5})\bigr) \exp\!\left(-\sum_{n\in A} \frac{2\pi^2p_n(1-p_n)h^2}{n^2}\right),$$

where the error may be complex. For large $N$ its real part is at least $-1/6$, and the exponential is positive. Hence

$$\frac1Q\sum_{|h|\leq H}\operatorname{Re}F(h) \geq\frac5{6Q}\sum_{|h|\leq H} \exp\!\left(-\sum_{n\in A} \frac{2\pi^2p_n(1-p_n)h^2}{n^2}\right) \geq\frac5{6Q},$$

using just the term $h=0$ for the last inequality.

For the remaining frequencies $H<|h|\leq M/2$, we have $|h|/n\leq1/2$. The absolute-value bound in Fact 2.5 gives

$$|F(h)|\leq\prod_{n\in A} \left(1-\frac{8p_n(1-p_n)h^2}{n^2}\right) \leq\exp\!\left(-8h^2\sum_{n\in A} \frac{p_n(1-p_n)}{n^2}\right).$$

Put $\delta=(\log N)^{-2}$. For large $N$, $p_n(1-p_n)\geq\delta(1-\delta)\geq\delta/2$. Since $n\leq N$, $|A|\geq N^{0.95}$, and $M\geq N^{0.95}$,

$$|F(h)| \leq\exp\!\left(-\frac{4\delta |A|M^{6/5}}{N^2}\right) \leq\exp\!\left(-\frac{4N^{0.09}}{(\log N)^2}\right).$$

There are at most $M+1\leq N+1$ integer frequencies in this range. Their total absolute contribution is consequently $o(1/Q)$, and for large $N$ is at most $1/(12Q)$. Combining the two ranges gives $5/(6Q)-1/(12Q)=3/(4Q)$, as claimed.

Source formula corrections. The source propagates Fact 2.5's $2\pi$ coefficient typo; the proof above uses $2\pi^2$, as required by the displayed Taylor expansion. Its first proof display also uses $\exp(p_nh/n)$ where the cancellation requires the phase $e(p_nh/n)$. The final tail estimate here retains the factor $1/Q$, making its comparison with the major-arc lower bound explicit. These are local formula clarifications; the argument and constants in the conclusion are those of the source.

Dependencies. Fact 2.5, the summability of $n^{-3}$, and elementary exponential estimates.

Bears on. #298 and #299, through the major-arc part of the quantitative reciprocal-sum criterion.