$n^2$ times an equidistributed sequence

Question

Suppose you have an equidistributed sequence $\{\alpha_n\}_{n=1}^\infty\subseteq(0,1)$, does $n^2\alpha_n\to\infty$ always? I don't know how to even start. Is it something like "it diverges for almost every sequence"? I am almost sure that $n\alpha_n$ isn't bounded from below by anything other than $0$.

What if the $\alpha$'s are equidistributed but in a "deterministic" way like $\alpha_n=\sqrt{2}\,n\mod 1$. In this case, does the divergence depend on the number and how is it approximated by rationals?

Thanks

Answer 1

What do you mean when you say $n^2\alpha_n \to \infty$? There is one particular interpretation of such a 'limit':

When $f(n) \to \infty$, we mean that for every $L$, there exists a $k$ such that for all $n > k$ we have $f(n) > L$.

We have $P(n^2\alpha_n > L) = 1 - Ln^{-2}$ and

$$P(\forall n > k: n^2\alpha_n > L) = \prod_{n=k+1}^\infty \left(1 - Ln^{-2}\right)$$

The issue is, for any $L$, there must exist a $k$ (and it must thus be a finite integer). But as close as the above product approaches $1$, we know just from the first term that:

$$P(\forall n > k: n^2\alpha_n > L) <1 - \frac{L}{(k+1)^2} < 1$$

So for any choice of $L$, no matter how big $k$ is, we end up with a non-zero probability that our sequence does not meet the requirements, and thus we can not prove that $n^2\alpha_n \to \infty$.