Summation Rules/Dividing Sums

Question

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. — user328442, Nov 20 '17 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 '17 at 18:16

user328442 · Accepted Answer · 2017-11-20T18:23:42.733

1

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:

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

edited Nov 20 '17 at 18:23

answered Nov 20 '17 at 18:16

user328442

2,705

Thank you for your answer and yes there was a small mistake in my post. – PAS Nov 21 '17 at 11:28

Summation Rules/Dividing Sums

1 Answers1