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][1] of this theorem on Terry Tao's blog (see Corollary's 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][2]. (**UPDATE:** George Lowther, below, suggests that Gowers may have a typo.) [Wikipedia][3] 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][4] over at math.SE. So any useful comments you have on that question would be appreciated as well.


  [1]: http://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/#vdc-eq
  [2]: http://www.dpmms.cam.ac.uk/~wtg10/addnoth.notes.ps
  [3]: http://en.wikipedia.org/wiki/Weyl%2527s_inequality
  [4]: http://math.stackexchange.com/questions/2270/convergence-of-sum-n1-infty-sinnk-n