(related to Proposition: Basic Calculations Involving Indefinite Sums)
- Chose the antiderivatives $F,G:\mathbb C\to\mathbb C$ of $f,g$, i.e. $\Delta F(x)=f(x)$ and $\Delta G(x)=g(x)$.
- Applying basic calculations involving the difference operator (no. 2), we get $\Delta (F(x)\pm G(x))=\Delta F(x)\pm \Delta G(x).$
- Summing on both sides yields $(F(x)\pm G(x))=F(x)\pm G(x).$
- This means $\sum (f(x)\pm g(x))=\sum f(x)\pm \sum g(x).$
- The product rule of the difference operator, can be re-written for suitable complex-valued functions $f,g$ as follows: $g(x)\Delta f(x)=\Delta (fg)(x)-f(x+1)\Delta g(x).$
- Summing on both sides yields $\sum g(x)\Delta f(x)=f(x)g(x)-\sum f(x+1)\Delta g(x).$