Why does this relation hold? Representing sum as (d-1) derivative of a function

Question

I'm having trouble undestanding the first equality (for starters). Can someone help - maybe refer me to some of the theory that is behind this.

Screenshot from "Sparse Grids in a Nutshell" by Jochen Garcke.

Answer 1

Maybe it's easier to understand backwards: Taking the $(d-1)$th derivative gives $$ \frac{\mathrm{d}^{(d-1)}}{\mathrm{d}x^{(d-1)}} \left[ \sum_{i=0}^{\infty} x^{i+n+d-1} \right] = \sum_{i=0}^{\infty} \left[ \frac{\mathrm{d}^{(d-1)}}{\mathrm{d}x^{(d-1)}} x^{i+n+d-1} \right] = \sum_{i=0}^{\infty} \frac{(i+n+d-1)!}{(i+n)!} x^{i+n}, $$ since each of the $(d-1)$ differentiations pulls down the power of $x$ and subsequently lowers the power by 1. Multiplying the above result by the prefactor $\frac{x^{-n}}{(d-1)!}$ the leads to the first equality in the equation you posted.

The second step can be seen by using the geometric series, i.e. $\sum_{i=0}^{\infty} x^i = \frac{1}{1 - x}$, which holds for $|x| < 1$.