Badly Approximable Numbers: Lower Bound for Sum of Reciprocals of Fractional Parts

Question

I am having a bit of trouble proving the following fact that I have seen in a couple of references on Diophantine approximation.

Let $||x|| = \min\{|x - p| ~|~ p \in \mathbb{Z}\}$. Recall that a number $\alpha$ is badly approximable if there exists $C > 0$ such that $||n \alpha|| > C/n$ for all $n \in \mathbb{N}$. Equivalently, $\alpha$ is badly approximably if it has bounded partial quotients. That is, if $\alpha = [a_0, a_1, \ldots]$, then there exists $A > 0$ such that $a_i < A$ for all $i$.

Let $\alpha$ be a badly approximable irrational number, and let $\{x\} = x - \lfloor x \rfloor$ be the fractional part function. There exists $c > 0$ such that for sufficiently large $N$,

$$ \sum_{n = 1}^{N} \frac{1}{\{n \alpha\}} \geq c N \log(N). $$

Could anyone offer a hint as to how to prove this?

Answer 1

I figured out the following proof.

Recall that $p/q$ (with $p,q$ relatively prime) is called a 1-convergent of $\alpha$ if $|q\alpha - p| < 1/q$.

Lemma. If $p/q$ is a 1-convergent of $\alpha$, then

$$ \sum_{n=1}^{q} \frac{1}{\{n\alpha\}} > \frac{q}{2} \log(q). $$

Proof. If $n < q$, then

$$ \left|n\alpha - n \frac{p}{q}\right| = \frac{n}{q} |q\alpha - p| < \frac{n}{q^2} < \frac{1}{q}. $$

It follows that

$$ n\alpha = n \frac{p}{q} + \frac{\theta}{q}, $$

where $|\theta| < 1$. Since $p$ is relatively prime to $q$ and $n < q$, $\{np/q\} \geq 1/q$, so that $\lfloor n\alpha \rfloor = \lfloor np/q \rfloor$. It follows that

$$ \{n\alpha\} < \left\{n \frac{p}{q}\right\} + \frac{1}{q} \leq 2\left\{n \frac{p}{q}\right\}. $$

Hence,

$$ \sum_{n=1}^{q-1} \frac{1}{\{n\alpha\}} \geq \sum_{n=1}^{q-1} \frac{1}{2\{np/q\}} = \frac{q}{2} \sum_{k=1}^{q-1} \frac{1}{k}. $$

(The inequality is not strict because it is possible that $q = 1$, so that both sides are trivially zero.) Since obviously $\{q\alpha\} < 2$,

$$ \sum_{n=1}^{q} \frac{1}{\{n\alpha\}} > \frac{q}{2} \sum_{k=1}^{q-1} \frac{1}{k} + \frac12 = \frac{q}{2} \sum_{k=1}^{q} \frac{1}{k} > \frac{q}{2} \log(q). $$

$\square$

Theorem. Suppose $\alpha$ is badly approximable. Then there exists $c > 0$ such that for sufficiently large $N$,

$$ \sum_{n=1}^{N} \frac{1}{\{n\alpha\}} > cN \log(N). $$

Proof. Since $\alpha$ is badly approximable, it is of constant type. (See theorem 6 of chapter 2 of Lang's Introduction to Diophantine Approximations.) That is, there exists $c' > 1$ such that for every sufficiently large $N$, there exists a 1-convergent $p/q$ such that $N/c' \leq q < N$. For sufficiently large $N$, pick such a $q$. Then

$$ \sum_{n=1}^{N} \frac{1}{\{n\alpha\}} > \sum_{n=1}^{q} \frac{1}{\{n\alpha\}}. $$

Using the lemma,

$$ \sum_{n=1}^{N} \frac{1}{\{n\alpha\}} > \frac{q}{2} \log(q) \geq \frac{N}{2c'} \log\left(\frac{N}{c'}\right). $$

Choose arbitrary $c'' \in (0,1)$. Then for sufficiently large $N$,

$$ (1 - c'') N \log(N) > \log(c') N, $$

so that

$$ c'' N \log(N) < N \log\left(\frac{N}{c'}\right). $$

Let $c = c''/2c'$. Then for sufficiently large $N$,

$$ \sum_{n=1}^{N} \frac{1}{\{n\alpha\}} > cN \log(N). $$

$\square$