Real and positive sum of complex number in the unit circle Let $\{x_i\}_{i=1}^n$ be a collection of complex numbers such that $|x_i| \leq 1\:\forall i = 1,...,n$, $|x_1|=1$ and
$$\sum_{i=1}^n x_i^k > 0 $$
for all positive integers $k$. Ie. this sum is real and strictly positive.
I am trying to prove that $\limsup_{k\rightarrow \infty} (\sum_{i=1}^n x_i^k)^{1/k} = 1$. But I don't really know how to go about it...
I know already that if there is a $-1$ in the collection then the $\liminf$ does not go to $1$. But any other example that I tried seems to go asymptotically to $1$ for both $\liminf$ and $\limsup$. 
I also tried to reformulate the problem, it is equivalent to proving:
Let $\{x_k\}_{i=1}^n$ be a collection of complex numbers such that $|x_i| = 1\:\forall i = 1,...,n$ and 
$$\sum_{i=1}^n x_i^k \geq 0 .$$
Prove that $\limsup_{k\rightarrow \infty} (\sum_{i=1}^n x_i^k)^{1/k} = 1$
In this version I noticed that if an element of the collection is not real, then also its conjugate will be in the collection, because of the fact that the sum has to be real, but as it has to be positive then there have to be at least 2 ones in the collection. But the thing is that I don't know how to prove these last observations...
I'd be glad if someone could help. If I formulated anything in a too vague way, please let me know and I'll clarify.
 A: Let $a_k =\sum_{i=1}^n x_i^k >0, k \ge 1$. Note that $a_k \le n$ so $\limsup_{k\rightarrow \infty} (a_k)^{1/k} \le 1$ and note also that $\sum_{k \ge 1}a_k = \infty$ since for any $x_j \ne 1$ the sum in $j$ is finite and we have $x_1=1$ and now the conclusion follows trivially since if $\limsup_{k\rightarrow \infty} (a_k)^{1/k} = b <1$, $a_k \le (\frac{1+b}{2})^k, k \ge k_0$ and $\sum{a_k}$ finite
Note that this solution applied to an earlier post where $x_1=1$, but we can adapt it as follows: let $x_1,..x_m$ the numbers among $x_j$ that have absolute value $1$ and note that $m \ge 1$ and then let $\theta_j \in (-\pi, \pi]$ their arguments and $\alpha_j=\frac{\theta_j}{2\pi}$. If all $\alpha_j$ are rational then obviously for all $q$ mutliple of the lcm of the denominators of $\alpha_j$, we have $x_j^q=1, j=1,..,m$ so $a_q=m-$small, for all such $q$ large enough and hence $\sum{a_k} =\infty$ and we are done.
If not all $\alpha_j, j=1,..m$ are rational, applying the Dirichlet approximation theorem with $N$ arbitrary large, gives us integers $p_j(N),q(N), |q(N)\theta_j - 2\pi p_j(N)| \le \frac{2\pi}{N^{\frac{1}{m}}}$  and $q(N) \to \infty$ with $N$ since not all $\alpha_j$ are rational hence not all $\theta_j - 2\pi \frac{p_j(N)}{q(N)}, j=1,..m$ can be zero, so as $N^{\frac{1}{m}} \to \infty$, $q(N)$ must also go to infinity. But then again it follows $a_{q(N)} \ge m -$ small for $N$ large enough, so again $\sum{a_k} =\infty$
A: (Using the answer to this question) + Cauchy Hadamard theorem

We can define the function $$f(z)= \sum_{i=1}^n \frac{1}{1-x_i\cdot z}$$
This is a meromorphic function having simple poles at $\frac{1}{x_i}, 1\leq 1\leq n$.
The Taylor series of $f$ with centre $0$, $$\sum_{i=1}^n\sum_{k\geq 0} x_i^kz^k =  \sum_{k\geq 0}\left(\sum_{i=1}^n x_i^k\right)z^k$$
Therefore the radius of convergence $$R = \min_{1\leq i\leq n}\frac{1}{|x_i|}=  1/\max_{1\leq i\leq n}|x_i|$$

This last line follows from the fact that $f$ is meromorphic and the poles are known. Since in this case $\max |x_i| = 1$ we have $R=1$. Therefore by the Cauchy Hadamard Theorem
$$\limsup_{k\rightarrow\infty} \left(\sum_{i=1}^n x_i^k\right)^{1/k} = \limsup_{k\rightarrow\infty} \left|\sum_{i=1}^n x_i^k\right|^{1/k} = 1/R = 1 $$
