Let $\alpha_1, \alpha_2, \dots$ be an infinite sequence of real numbers such that any finite subset is linearly independent over $\mathbb{Q}$. Let $f(N)$ be the number of tuples $(m_1, \dots, m_N)$ with $m_i \in \mathbb{Z}$, $|m_i| \leq N$. for which $$\left|\sum_{i=1}^{N} m_i \alpha_i \right|< \frac{1}{N^{100}}.$$

(where 100 could be your favorite constant). Is there any known way to put an asymptotic upper bound on $f(N)$?

For concreteness (or if it makes the question possible to answer!) you could take the $\alpha_i$ to be the square roots of square free integers (ordered however you like, although I do have a particular ordering in mind). If that doesn't work, any effectively computable $\alpha_i$ would be welcome.

It seems that one might be able to say something of the sort using Evertse's quantitative subspace theorem but it isn't really clear to me how. Any pointers (or even conjectures) here would be greatly appreciated!

Let $\alpha_1, \alpha_2, \dots$ be an infinite sequence of real numbers such that any finite subset is linearly independent over $\mathbb{Q}$. Let $f(N)$ be the number of tuples $(m_1, \dots, m_N)$ with $m_i \in \mathbb{Z}$, $|m_i| \leq N$. for which $$\left|\sum_{i=1}^{N} m_i \alpha_i \right|< \frac{1}{N^{100}}.$$

(where 100 could be your favorite constant, or maybe something sublinear if that's not possible). Is there any known way to put an asymptotic upper bound on $f(N)$ better than the trivial one of $(2N+1)^N$?

For concreteness (or if it makes the question possible to answer!) you could take the $\alpha_i$ to be the square roots of square free integers (ordered however you like, although I do have a particular ordering in mind). If that doesn't work, any effectively computable $\alpha_i$ would be welcome.

It seems that one might be able to say something of the sort using Evertse's quantitative subspace theorem but it isn't really clear to me how. Any pointers (or even conjectures) here would be greatly appreciated!

EDIT: I just noticed something, though it's not super useful: If one takes $\alpha_1 = 1$, $\alpha_k = \log p_k$ ($p_k$ the $k$th prime), then for large $N$, for each $i$ it cannot be the case that just $m_1$ and $m_i$ are non-zero by a quantitative Baker's theorem. So we have at least $f(N) \leq (2N + 1)^N - (N-1)(2N+1)^2$. This isn't super useful for me since this has the same asymptotics as the trivial bound, but might be helpful to think about.

Let $\alpha_1, \alpha_2, \dots$ be an infinite sequence of real numbers such that any finite subset is linearly independent over $\mathbb{Q}$. Let $f(N)$ be the number of tuples $(m_1, \dots, m_N)$ with $m_i \in \mathbb{Z}$, $|m_i| \leq N$. for which $$\left|\sum_{i=1}^{N} m_i \alpha_i \right|< \frac{1}{N^{100}}.$$

(where 100 could be your favorite constant). Is there any known way to put an asymptotic upper bound on $f(N)$?

It seems that one might be able to say something of the sort using Evertse's quantitative subspace theorem but it isn't really clear to me how. Any pointers (or even conjectures) here would be greatly appreciated!

Let $\alpha_1, \alpha_2, \dots$ be an infinite sequence of real numbers such that any finite subset is linearly independent over $\mathbb{Q}$. Let $f(N)$ be the number of tuples $(m_1, \dots, m_N)$ with $m_i \in \mathbb{Z}$, $|m_i| \leq N$. for which $$\left|\sum_{i=1}^{N} m_i \alpha_i \right|< \frac{1}{N^{100}}.$$

(where 100 could be your favorite constant). Is there any known way to put an asymptotic upper bound on $f(N)$?

For concreteness (or if it makes the question possible to answer!) you could take the $\alpha_i$ to be the square roots of square free integers (ordered however you like, although I do have a particular ordering in mind). If that doesn't work, any effectively computable $\alpha_i$ would be welcome.

It seems that one might be able to say something of the sort using Evertse's quantitative subspace theorem but it isn't really clear to me how. Any pointers (or even conjectures) here would be greatly appreciated!