 Solution

Write $e(t)=e^{2\pi it}$ and $N=n+1$. A convenient form of the quadratic Weyl inequality is the following: if $q\ge 1$, $(a,q)=1$ and $|\alpha -a/q|\le q^{-2}$, thenThe implied constant is absolute. In particular the familiar square-root saving is $|S|\ll (N^2/q+N+q{)}^{1/2}\log (2q{)}^{1/2}$. We prove all estimates used in this statement.

First, the finite geometric series identity giveswhere $\Vert \theta \Vert$ denotes distance to the nearest integer and the second bound is interpreted as infinite when that distance is zero. Indeed the sum is $(1-e(L\theta ))/(1-e(\theta ))$, and $|1-e(\theta )|=2|\sin \pi \theta |\ge 4\Vert \theta \Vert$. This proves the exponential geometric sum bound.

Expand the squared quadratic exponential sum by the difference $h$ between its indices. Since $(x+h{)}^2-x^2=2hx+h^2$, the diagonal and the paired off-diagonals giveFor $q\ge 16$, partition the $h$ into consecutive blocks of length at most $L=\lfloor q/8\rfloor$. Within one block, two distinct indices have difference $0<|d|<q/2$. Since the denominator of $2a/q$ after reduction is at least $q/2$, $\Vert 2ad/q\Vert \ge 1/q$. ThereforeThus the points $2\alpha h$ in a block are separated on the circle by at least $3/(4q)$.

For any such separated set, at most an absolute number of points lie in each pair of arcs whose distances from zero are between $j/(4q)$ and $(j+1)/(4q)$. The closest few contribute $O(N)$ to the capped reciprocal sum; the remaining arcs contribute $O(q/j)$ for $1\le j\le 4q$. Summing the harmonic series, using ${\sum }_{j\le m}{j}^{-1}\le 1+\log m$, proves a block bound $O(N+q\log (2q))$. This is the required separated reciprocal-distance sum estimate, with its proof supplied here. There are $O(N/q+1)$ blocks, soFor $q<16$, the trivial bound $|S|\le N$ already implies this estimate after increasing the absolute constant. This completes the proof, including the small denominators and the possible common factor of $2$ and $q$.