Suppose that $x_j$ is an $n$-periodic sequence. Show that $$\sum_{j=m}^{m+n-1}x_j=\sum_{j=0}^{n-1}x_j.$$
So far I have tried playing around with the indices of the sequence and have \begin{align*} \sum_{j=m}^{m+n-1}x_j &= \sum_{j=m}^{m+n}x_j-x_{m+n} \\ &= \sum_{j=m}^{m+n}x_j-x_m \\ &= \sum_{j=m+1}^{m+n}x_j \end{align*} but I don't know what to make of this. I know the continuous equivalent statement for the integral of continuous functions holds but I don't know how to show the discrete case.
What you have shown is for any $m$ the sum over the range $R=[m,m+n-1]$ is the same as the sum over the range $S=[m+1,m+n]$ which is the range of $R$ with each entry increased by $1.$ So you can finish, since the original sum on the right of your desired equation is over the range $[0,n-1]$ and by repeatedly adding 1 to this range you arrive after $m$ steps at the range $[m,m+n-1]$ of the sum on the left.