Simultaneous rational approximation to transcendental and algebraically independent numbers

I’m interested in the problem of simultaneous rational approximation to $k\geq 2$ numbers $\alpha_1,…,\alpha_k$ in the generic case where the $\alpha_j$ are transcendental and algebraically independent. More precisely, what is the largest $m$ such that there exists an infinite sequence of positive integer denominators $q$ and corresponding integers $n_j$ (depending on q) such that

$|\alpha_j-n_j/q|<1/q^{m+1/k}\qquad (1\leq j\leq k)$

Alternatively, what is the largest $m$ such that the above holds for except on a set of $k$-tuples $(\alpha_1,\ldots,\alpha_k)$ of measure 0. The classical Dirichlet theorem shows $m\geq 1$ (with no exceptions). But is the generic value $m\geq 2$? I'm hoping that this is well known among number theorists; any pertinent references would be much appreciated.

• FWIW, on MathOverflow, 'Title' is meant to describe your question and not your job title :) – Glorfindel Jul 7 '18 at 16:39

For any $m>1$ the set of such $k$-tuples $\alpha:=(\alpha_1,\dots,\alpha_k)$ have measure zero. For seeing this restrict onto $\alpha\in [0,1]^k$ and denote by $\Omega_q$ the set of $\alpha$'s in $[0,1]^k$ for which this $q$ works. The measure of $\Omega_q$ does not exceed, say, $2^k q^{k(1-m-1/k)}=2^k q^{-1-k(m-1)}$. Thus the sum of measures of $\Omega_q$'s is finite and by (easy part of) Borel -- Cantelli lemma almost any point $\alpha\in [0,1]^k$ belongs only to finitely many $\Omega_q$'s.