Let $x\_1,x\_2,...,x\_k$ be irrational number,is it always true that:
$\liminf\_{n\rightarrow\infty} \sum\_{i=1}^k (nx\_i)=0$ (where $(x)$ denotes the fractional part of $x$)


If not,what are the necessary and sufficient conditions that {$x\_i$} must satisfy so that $\liminf\_{n\rightarrow\infty} \sum\_{i=1}^k (nx\_i)=0$ 



sum of fractional parts (nx_i),x_i are irrational
Here is a counterexample: $x\_1 = \sqrt2, x\_2 = 1-\sqrt2$.
For $n>0$, $(nx\_1) + (n x\_2) =1$ so the limit infimum is $1$, not $0$.


I think the other answers assumed that you meant $x-[x]$, where $[x]$ is the nearest integer to $x$, instead of $(x)$. 


If there is no rational dependency, then the multiples are dense in the torus $(\mathbb{R}/\mathbb{Z})^k$, so the limit infimum is $0$. If there is some rational dependency, there will always be small multiples (close to the origin in $(\mathbb{R}/\mathbb{Z})^k$), but these might not be positive.


Consider the closure of multiples of $\langle x\_1,...,x\_k \rangle$ in the torus $(\mathbb{R}/\mathbb{Z})^k$. This is a closed subgroup, hence a finite union of parallel flat subtori. For $(x)$, the condition you want is that the tangent space to this subgroup at the origin intersects the positive orthant.


