Let $U$ be the infinite unitary group $\lim_{n\to\infty}U(n)$. It is well known that, over the rationals, $BU$ is homotopy equivalent to $\prod_{n=1}^\infty K(\mathbb{Z}, 2n)$.

**Question:** Is it true that, for any $k\geq 0$, the homotopy group $\pi_{2k+1}(BU(m))$ is 0 for $m$ sufficiently large?