How to prove: for any $m \in Z$, $\sum_{n=m}^{m+N-1} h(n) =\sum_{n=0}^{N-1} h(n)$

Question

Suppose $h$ is a function on $Z$ that is periodic with period N, that is $h(n+N)=h(n)$ for all $n$. How to prove: for any $m \in Z$, $\sum_{n=m}^{m+N-1} h(n) =\sum_{n=0}^{N-1} h(n)$. In other words, any sum over an interval of length $N$ yields the same result. This is problem from the area of linear algebra and is a homework

Answer 1

Let $m \in \{0 \dots, N-1\}$. Then we have $$\sum_{n=m}^{m+N-1} h(n) = \sum_{n=m}^{N-1} h(n) + \sum_{n=0}^{m-1} h(n + N) = \sum_{n=m}^{N-1} h(n) + \sum_{n=0}^{m-1} h(n) = \sum_{n=0}^{N-1} h(n)$$ Now for $m \in \mathbb{Z}$ write $m = kN + r$ with $k \in \mathbb{Z}$ and $r \in \{0, \dots, N-1\}$. We clearly have $$\sum_{n=kN+r}^{k(N+1) + r -1} h(n) = \sum_{n=r}^{N+r-1} h(n+kN) = \sum_{n=r}^{N+r-1} h(n)$$ Now the last expression equals $\sum_{n=0}^{N-1} h(n)$ by the first part of the argument.

Answer 2

HINT:

Expand $$\sum_{n=m}^{m+N-1}h(n)$$ What do you know about $h$?

Answer 3

The important point here is that any set of $N$ consecutive integers is a complete residue system modulo $N$. See complete residue system. We consider the index modulo $N$ since $h(n)$ has period $N.$ i.e. the numbers $ \lbrace m,m+1,m+2,\ldots,m+N-1 \rbrace $ modulo $N$ are $ \lbrace 0,1,2,\ldots,N-1 \rbrace $ in some order.

Your identity follows immediately from this fact.