I am working with generating functions and am required to prove that the generating function for the sequence $\{a_k\}$ where $a_k = (-8)^k$ for all integers $k\geq0$ is $\cfrac{1}{1+8x}$ and I have gotten it to:
$$g(x)[1+8x] = \lim \limits_{n\to\infty} 1 - (-8)^{n+1}x^{n+1}$$
I can see that I want to take $(-8)^{n+1}x^{n+1}\to0$, but I don't know how I can say this is alright.
It works for $0\lt x\lt\lt\lt1$ I suppose, but I don't see how I can justify this.
This has been shown to me. It is a very similar question, and they have gotten to teh stage where $g(x)-xg(x)=1$, this is what I want, but this ugly one above can only reduce if $x\lt\frac18$. The current only answer says that I can't assume this, and I feel as though that is true. But his formula assumes what he is saying I can't.
When using generating functions we treat the series $$\sum_{n=0}^{\infty}a_nx^n$$ as formal expression, and we dont care about convergence rate and stuff like that.
The notation $\frac{1}{1+8x}$ is nothing but a simple notation, and we dont actually care about subtituting x with anything, just with the cofficient of $x^k$ in the series.
For your series the generating function is $$f(x) = \sum_{n=0}^{\infty}(-8)^nx^n = \sum_{n=0}^{\infty}(-8x)^n = \frac{1}{1+8x}$$
Where the last part is that notation that is convinient since its the sum of the geometric series