[Math] Summation Rules/Dividing Sums

summation

Is there a way to simplify the following expression?

$${\sum_{i=1}^n (1-g(x))^{i-1}} \over {\sum_{i=1}^n (1-g(x))^i} $$

With $0<g(x)<1$ and $n \in \mathbb{N} $. This expression is the same as:

$${\sum_{i=0}^{n-1} (1-g(x))^{i}} \over {\sum_{i=1}^n (1-g(x))^i} $$

Let's say $(1-g(x))=h(x)$. I can rewrite the above expression as:

$${1-h(x)^n\over {1-h(x)}}\over {{h(x)^{n+1}-h(x)} \over {1-h(x)}} $$

$${1-h(x)^n}\over {h(x)^{n+1}-h(x)} $$

$${1-h(x)^n}\over {h(x)(h(x)^n-1) } $$

But this does not really help me to make it any clearer for me.

Is there another simplification I'm missing?

Best Answer

Found a small mistake, note that $$\sum_{k=1}^n x^k = \frac{x^{n+1}-x}{x-1}$$ and so we have a sign error. To fix this, just tack on a negative sign. This leads to:

$$\frac{h(x)^n-1}{h(x)^{n+1}-h(x)} = \frac{1}{h(x)} \cdot \frac{h(x)^n-1}{h(x)^n-1} = \frac{1}{h(x)} = \frac{1}{1-g(x)}$$

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