show/hide this revision's text 4 edited in light of Benoit's answer

After writing out

I made the following failed attempt , I agree. to prove the result using Weyl's inequality, which failed. It only manages to show that $\liminf S_N/N\to0$. However, as writtenmentioned in Benoît's answer, does not imply with a little bit more work the equidistribution theorem does follow.

I'll leave

The problem with the following failed attempt here, and you can see proof below is that it fails precisely when the convergents $q_i$ can grow so fast that they jump right over the (huge) intervals in (1) below - as David mentions in the question.It only shows that $\liminf S_N/N\to0$, unless there is a bound on how fast the convergents grow.

\epsilon^{-1}(\log begin{array}{}\displaystyle\epsilon^{-1}(\log N)^{d(1-2^{-d})}\le q\le \epsilon N^d/(\log N)^{d(1-2^{-d})}.&&(1)\epsilon begin{array}{}\displaystyle\epsilon N^d/(\log N)^{d(1-2^{-d})}>\epsilon^{-1}(\log(N+1))^{d(1-2^{-d})}.&&(2)Choosing Choose $N_0$ large enough that, for all $N\ge N_0$, this inequality is satisfied and $(\log N)^{d(1-2^{-d})}/N<\epsilon$. <\epsilon$ and (2) is satisfied. Then, as the union of the intervals (1) over $N\ge N_0$ covers the range $[\epsilon^{-1}(\log N_0)^{d(1-2^{-d})},\infty)$.As there will be exist infinitely many coprime p,q with $\vert\theta-p/q\vert\le1/q^2$, we can take $q>\epsilon^{-1}(\log N_0)^{d(1-2^{-d})}$satisfying .

Initially, I attempted to go from here to conclude that $S_N/N$ is small for $N\ge N_0$, which was the requirementserror in my first version of this answer. Instead, this argument just shows that q must lie in one of the intervals (1) for some $N\ge N_0$, in which case $S_N/N\le 100(3\epsilon)^{1/(2^d-1)}$for $N>N_0$, so . So, we can find large N making $S_N/N$ becomes arbitrarily as small for large Nas we like, and $\liminf S_N/N\to0$.
show/hide this revision's text 3 oops, point out the flaw

Yes

After writing out the following failed attempt, it I agree. Weyl's inequality, as written, does seem to not imply the equidistribution theorem.Hopefully there's no mistakes in

I'll leave the following . (Edit: think I may have messed up failed attempt here, and you can see that it fails precisely when the convergents $q_i$ grow so fast that they jump right over the (huge) intervals below - as David mentions in the question. This just It only shows that $\liminf S_N/N\to0$)S_N/N\to0$, unless there is a bound on how fast the convergents grow.

The inequality you wrote is equivalent to $$ S_N/N\le100\left(\frac{(\log N)^{d(1-2^{-d})}}{q}+\frac{(\log N)^{d(1-2^{-d})}}{N}+\frac{(\log N)^{d(1-2^{-d})}q}{N^d}\right)^{1/(2^d-1)} $$ As long as it can be shown that each of the three terms inside the parentheses is less than or equal to any fixed $\epsilon>0$ for large enough N, then $S_N/N\to0$. The middle term tends to zero, so that is no problem. For the remaining two terms to be less than $\epsilon$, we need $$ \epsilon^{-1}(\log N)^{d(1-2^{-d})}\le q\le \epsilon N^d/(\log N)^{d(1-2^{-d})}. $$ For large N, this gives a big range from which to choose our q. In fact, as you increase N these intervals eventually overlap $$ \epsilon N^d/(\log N)^{d(1-2^{-d})}>\epsilon^{-1}(\log(N+1))^{d(1-2^{-d})}. $$ Choosing $N_0$ large enough that, for all $N\ge N_0$, this inequality is satisfied and $(\log N)^{d(1-2^{-d})}/N<\epsilon$. Then, as there will be $q>\epsilon^{-1}(\log N_0)^{d(1-2^{-d})}$ satisfying the requirements, this shows that $S_N/N\le 100(3\epsilon)^{1/(2^d-1)}$ for $N>N_0$, so $S_N/N$ becomes arbitrarily small for large N.
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Yes, it does seem to imply the equidistribution theorem. Hopefully there's no mistakes in the following. (Edit: think I may have messed up here. This just shows that $\liminf S_N/N\to0$).

The inequality you wrote is equivalent to $$ S_N/N\le100\left(\frac{(\log N)^{d(1-2^{-d})}}{q}+\frac{(\log N)^{d(1-2^{-d})}}{N}+\frac{(\log N)^{d(1-2^{-d})}q}{N^d}\right)^{1/(2^d-1)} $$ As long as it can be shown that each of the three terms inside the parentheses is less than or equal to any fixed $\epsilon>0$ for large enough N, then $S_N/N\to0$. The middle term tends to zero, so that is no problem. For the remaining two terms to be less than $\epsilon$, we need $$ \epsilon^{-1}(\log N)^{d(1-2^{-d})}\le q\le \epsilon N^d/(\log N)^{d(1-2^{-d})}. $$ For large N, this gives a big range from which to choose our q. In fact, as you increase N these intervals eventually overlap $$ \epsilon N^d/(\log N)^{d(1-2^{-d})}>\epsilon^{-1}(\log(N+1))^{d(1-2^{-d})}. $$ Choosing $N_0$ large enough that, for all $N\ge N_0$, this inequality is satisfied and $(\log N)^{d(1-2^{-d})}/N<\epsilon$. Then, as there will be $q>\epsilon^{-1}(\log N_0)^{d(1-2^{-d})}$ satisfying the requirements, this shows that $S_N/N\le 100(3\epsilon)^{1/(2^d-1)}$ for $N>N_0$, so $S_N/N$ becomes arbitrarily small for large N.
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