Question

What is the geometric power series?

Answer 1

`a+ax+ax^{2}+ax^{3}+cdots`, which converges to `a/(1-x)` when `|x|<1`. More generally, you could also write `a+a(x-c)+a(x-c)^{2}+a(x-c)^{3}+cdots`, which converges to `a/(1-(x-c))=a/((1+c)-x)` when `|x-c| < 1`.