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For each $t \in \mathbf{R}$, let $\mathrm{frac}(t)$ be its fractional part.

Question. Fix reals $\alpha_1,\ldots,\alpha_k \in (0,1)$ such that $\sum_{i\le k}\alpha_i<1$. Do there exist infinitely many positive integers $n$ such that $$ \sum_{i\le k}\mathrm{frac}(n\alpha_i)<1\,\,\,\,? $$

The answer is affirmative if $\{\alpha_1,\ldots,\alpha_k\}\setminus \mathbf{Q}$ is a set which is linearly independent over $\mathbf{Q}$ (applying the multidimensional equidistribution theorem). But I don't know in general.

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    $\begingroup$ Yes, because you can make $n\alpha_i$ very close to integers simultaneously, so at the very next step you'll get the fractional parts almost the same as the original numbers. $\endgroup$
    – fedja
    Jan 29, 2020 at 16:47
  • $\begingroup$ Yes by Poincaré recurrence. $\endgroup$ Jan 29, 2020 at 16:48
  • $\begingroup$ @fedja Probably I am missing your point: also $0.99$ is close to an integer.. $\endgroup$ Jan 29, 2020 at 16:52
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    $\begingroup$ Yes, but $frac(0.99+0.23)$ is close to $0.23$ (next step means $n+1$ instead of $n$) $\endgroup$
    – fedja
    Jan 29, 2020 at 16:54

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