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 $\liminf_{n\rightarrow\infty} \sum_{i=1}^k (nx_i)=0$
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.
If you take any vector in Euclidean space, and look at its integer multiples modulo the integer lattice, we know what happens qualitatively. They are dense in a certain subtorus of the obvious torus. That's a version of Kronecker's theorem. Your question is about being near the origin infinitely often. So I say it's true.
Yes. This is the subject of "simultaneous diophantine approximation", googling which will produce many results, but you can take a look at Doug Hensley's notes:
http://www.math.tamu.edu/~Doug.Hensley/SimultaneousDiophantine.pdf
