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](http://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/#vdc-eq) 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](http://www.dpmms.cam.ac.uk/~wtg10/addnoth.notes.ps). (**UPDATE:** George Lowther, below, suggests that Gowers may have a typo.) [Wikipedia](http://en.wikipedia.org/wiki/Weyl%2527s_inequality) 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](https://math.stackexchange.com/questions/2270/convergence-of-sum-n1-infty-sinnk-n) over at math.SE. So any useful comments you have on that question would be appreciated as well.



Does Weyl's Inequality prove equidistribution?
I may be wrong, but it seems to me that in fact the liminf is sufficient since it is obtain via a bound that does not depend on $f$. I do not have time to check this in detail, so I apologize if this is all wrong. The idea is to use the points at which we have a good control on $S\_N$ (simultaneously for $f$ and all its translates, and it is the point that seems dubious to me) and then cut the summation interval into pieces of appropriate size.


Let us denote by $S\_N^t$ the exponential sum corresponding to the function $f(\cdot+t)$. Then if I am not mistaken, for all $\varepsilon$ there is a $N\_\varepsilon$ such that for all $t$, $S\_{N\_\varepsilon}^t\leqslant \varepsilon N\_\varepsilon$. Then given any $K$, one has 
$$S\_{KN\_\varepsilon}\leqslant \sum\_{k=1}^K S\_{N\_\varepsilon}^{kN\_\varepsilon}\leqslant K\varepsilon N\_\varepsilon.$$


Then, for all $N$, letting $K=\lfloor N/N\_\varepsilon \rfloor$ we get that $S\_N\leqslant K\varepsilon N\_\varepsilon + N\_\varepsilon\leqslant \varepsilon N+N\_\varepsilon$. It follows that $\limsup S\_N/N = 0$.


