Let $f(n) = \theta n^d + a_{d-1} n^{d-1} + \cdots a_1 n + a_0$ be a polynomial with real coefficients, and $\theta$ irrational. Let $S_N = \sum_{n=1}^N e^{2 \pi i f(n)}$. Weyl's Equidistribution theorem for polynomials is equivalent to the claim that $S_N/N \to 0$ as $N \to \infty$. You can read a nice proof of this theorem on Terry Tao's blog (see Corollaries 5 and 6). I had thought that Weyl's Inequality was supposed to be a more precise version of this bound. However, I can't actually figure out how to get Weyl's Inequality to imply the required claim!

Specifically, let $p/q$ be a rational number in lowest terms with $|\theta - p/q| \leq 1/q^2$. Weyl's Inequality is the bound:

$$S_N/N \leq 100 \left( \log N \right)^{d/2^d} \left( \frac{1}{q} + \frac{1}{N} + \frac{q}{N^d} \right)^{1/(2^d-1)}$$

Here are I am quoting from Timothy Gowers' notes. (UPDATE: George Lowther, below, suggests that Gowers may have a typo.) Wikipedia has a softer version, with more freedom in choosing parameters; I think my question applies to both versions.

Now, suppose that the convergents $p_i/q_i$ of $\theta$ grow so fast that $q_{i+1} > e^{(d+1) q_i}$. And take $N \approx e^{q_i}$. I get that, for any choice of $q$ with $|\theta - p/q| < 1/q^2$, either $1/q > 1/\log N$ or $q/N^d > 1$. This gives infinitely many $N$'s for which the right hand bound is useless (greater than $1$). So it seems that Weyl's inequality does not prove $S_N/N \to 0$.

Am I missing something?

The motivation for this question was my attempt to answer this question over at math.SE. So any useful comments you have on that question would be appreciated as well.