Vaughan Theorem 4.2: $|S(q,a)|\ll_{k,\varepsilon} q^{1-1/k+\varepsilon}$

Let $k\ge2$ and $\varepsilon>0$. There is a constant $C=C(k,\varepsilon)>0$ such that for every modulus $q\ge1$ and every integer $a$ with $(a,q)=1$,

This is the classical bound for complete exponential sums, Theorem 4.2 of Vaughan's The Hardy-Littlewood Method. It is sharper than the exponent $1-2^{1-k}+\varepsilon$ obtainable by applying Weyl's inequality at $N=q$: the two agree at $k=2$, and for every $k\ge3$ one has $1-\tfrac1k<1-2^{1-k}$, a saving that grows with $k$.

Structure of the proof. By the Chinese remainder theorem $q\mapsto S(q,a)$ is multiplicative in a twisted sense, $S(q_{1}q_{2},a)=S(q_{1},aq_{2}^{k-1})S(q_{2},aq_{1}^{k-1})$ for $(q_1,q_2)=1$, so everything reduces to prime powers, where $|S(p^{h},a)|\le k^{3}(p^{h})^{1-1/k}$. Multiplying these local estimates over the prime factorization gives

and the accumulated constant is absorbed by the divisor bound: choosing $J$ with $k^{3}\le2^{J}$ one has $(k^{3})^{\omega(q)}\le d(q)^{J}\ll_{\varepsilon}q^{\varepsilon}$. The $\varepsilon$ in the statement is exactly the price of this last step; the exponent $1-1/k$ itself is attained without loss.

Why it matters. In Waring's problem the local factors $A(q,n)$ are built from $S(q,a)$, and the absolute convergence of the singular series depends on how fast $|S(q,a)|$ decays below the trivial $q$. The exponent $1-1/k$ is what makes the singular series converge for $s>2k$ rather than only for $s$ exponentially large in $k$.