# Summation Rules/Dividing Sums

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?

• Start by cancelling out the $1-h(x)$ terms. After that, you can pull an $h(x)$ term from the denominator. Nov 20, 2017 at 18:13
• I edited the post but I still don't think this is much easier to use than the first expression.
– PAS
Nov 20, 2017 at 18:16
• $1-h(x)^n=-(h(x)^n-1)$ Nov 20, 2017 at 18:43

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:

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

• Thank you for your answer and yes there was a small mistake in my post.
– PAS
Nov 21, 2017 at 11:28