Tao Section 5: summing the Type I envelope over blocks

Summing the Type I pointwise envelope. Let $q\ge4$ with $(a,q)=1$, let $4\alpha=\frac aq+\beta$ with $|\beta|\le q^{-2}$, let $x>0$ and $U,V\ge40$ with $UV\le\frac x4$, and let $W$ be any nonnegative function on the positive odd integers $d\le UV$ satisfying the pointwise envelope

(the second alternative dropped when the sine vanishes). Then

This is the combinatorial half of the source's Type I estimate, separated from the analysis that produces the envelope. The argument splits the range of $d$: for $d\le\frac q2$ one has $\|4d\alpha\|_{\mathbb R/\mathbb Z}\ge\frac1q-\frac{q/2}{q^2}=\frac1{2q}$, hence $\frac1{|\sin(2\pi d\alpha)|}\le2q$, and the odd-restricted Vinogradov lemma bounds that contribution; for each subsequent block $2jq+\frac q2<d\le2(j+1)q+\frac q2$ the same lemma applies with the first alternative frozen at the left endpoint of the block, and the resulting harmonic sum over blocks is estimated by an integral test.

Why the additive 4 The integral test compares the decreasing summand with the integral over the preceding block of length $2q$, which covers the blocks $j\ge1$ but not $j=0$; the uncovered term is $\frac{x}{q/2}=\frac{x}{2q}\cdot4$. At $q=4$, $UV=40$ the sum $\sum_{0\le j\le4}\frac{x}{2\cdot4j+2}$ equals $0.7235x$ while $\frac x8\log24=0.3973x$. The statement above carries the restored term; the source's display omits it.

Formalization Note $W$ is an arbitrary nonnegative function on $\mathbb N$ constrained only on the positive odd $d\le UV$, so the statement is exactly the passage from the envelope to the two terms and carries no information about the exponential sums themselves. The index set is the platform's theorem51Divisors U V.