Power of a periodic sequence
The basic trick is to bound the series above and below. Let us do it on one side, for positive indices.  

For any $N&gt; 0$, you can write $N=kN_0+r_N$, with $0\le r_N&lt; N_0$. Then if $a_n$ (here $a_n = |x_n|^2$) is positive, $\sum_{n=0}^{N-1} a_n$ is increasing. Now  $kN_0 \le N&lt; (k+1)N_0$, hence you have:

$$ \sum_{n=0}^{kN_0-1} a_n  \le \sum_{n=0}^{N-1} a_n  \le   \sum_{n=0}^{(k+1)N_0-1} a_n   $$
or 

$$ k\sum_{n=0}^{N_0-1} a_n  \le \sum_{n=0}^{N-1} a_n  \le   (k+1)\sum_{n=0}^{N_0-1} a_n   $$

since  $a_n$ is  $N_0$-periodic. Now since 

$$\frac{1}{(k+1)N_0} \le \frac{1}{N} \le  \frac{1}{kN_0}$$
by a simple point-wise product of the threefold inequality (valid because positive):

$$ \frac{k}{(k+1)N_0} \sum_{n=0}^{N_0-1} a_n  \le \frac{1}{N}\sum_{n=0}^{N-1} a_n  \le    \frac{(k+1)}{kN_0}\sum_{n=0}^{N_0-1} a_n  \,. $$

Since $k/(k+1)\to 1$ and $(k+1)/k\to 1$ as $N\to\infty$, you have your result, mostly. You can split the original series in $\sum_{-N}^{N}a_n = \sum_{0}^{N} a_n+ \sum_{-N}^{0}a_n - a_0$,  take care of the positive and negative indices separately, and leave $a_0/(2N+1)$ tend to $0$.