Explanation of Note After Theorem 8.4

Let $h(n)$ be periodic with period $k$, i.e., $h(n+k)=h(n)$ for all integers $n$.

Let $k=5$ for illustration purpose.

$$\sum _{n=0}^{4}h(n)=\sum _{n=5}^{9}h(n)=\sum _{n=100}^{104}h(n)=\sum _{n=10,21,7,13,74}h(n).$$

$$\begin{array}{rl}\sum _{n=0}^{4}h(n)& =h(0)+h(1)+h(2)+h(3)+h(4).\\ \sum _{n=5}^{9}h(n)& \begin{array}{rl}& =h(5)+h(6)+h(7)+h(8)+h(9)\\ & =h(0)+h(1)+h(2)+h(3)+h(4).\end{array}\\ \sum _{n=10,21,7,13,74}^{\phantom{4}}h(n)& \begin{array}{rl}& =h(10)+h(21)+h(7)+h(13)+h(74)\\ & =h(0)+h(1)+h(2)+h(3)+h(4).\end{array}\end{array}$$