Power of a periodic sequence I am trying to find the way to reduce the standard expression to compute the power of a generic sequence $x(n)$:

$$P_{\text{x}}= \lim\limits_{N \to \infty}\frac{1}{2N + 1}\sum\limits_{n=-N}^{N}|x(n)|^2\text,$$

when $x(n) = x(n + N_0)$ ($N_0$-periodic sequence), into the simplified formula

$$P_{\text{x}}= \frac{1}{N_0}\sum\limits_{n=0}^{N_0-1}|x(n)|^2\text.$$

How can I demonstrate that second one could be obtained from the first one, using the hypothesis of periodicity? I've tried something like this, but I get stuck in after $\text{(iii)}$:

$$\begin{align}
N &amp;= N_0\tag{i}\\
P_{\text{x}}&amp;= \lim\limits_{N_0 \to \infty}\frac{1}{2N_0 + 1}\sum\limits_{n=-N_0}^{N_0}|x(n)|^2\\
&amp;= \lim\limits_{N_0 \to \infty}\frac{1}{2N_0 + 1}\sum\limits_{n=-N_0}^{N_0}|x(n + N_0)|^2\tag{ii}\\
&amp;m = n + N_0\text{, so }\\
P_{\text{x}}&amp;= \lim\limits_{N_0 \to \infty}\frac{1}{2N_0 + 1}\sum_{m=0}^{2N_0}|x(m)|^2\tag{iii}
\end{align}
$$
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 $0$.