# How to derive $\lim_{n\rightarrow ∞} \frac{1-r^{N+1}}{1-r}$

1. May 29, 2013

### Lebombo

How is $\lim_{n\rightarrow ∞} \frac{1-r^{N+1}}{1-r}$ derived?

Is it directly related to $S_{n}=\frac{a(1-r^{n})}{1-r}$?

If so, how does the $r^{n}$ become $r^{n+1}$

and how does the a disappear?

I've seen how to derive $S_{n}=\frac{a(1-r^{n})}{1-r}$, but have never come across $\lim_{n\rightarrow ∞} \frac{1-r^{N+1}}{1-r}$ until I just came across a brief mention of it in a video.

Any suggestions on where I could find a derivation/proof for it?

2. May 29, 2013

### Office_Shredder

Staff Emeritus
You have an N in the fraction but you're taking the limit for n. I'm assuming these are supposed to be the same thing.

There are several possibilities:

1) r > 1. Then rn gets really big as n goes to infinity, so this limit diverges
2) r = 1. Then the numerator and denominator are both zero and you shouldn't have written this down to begin with
3) r < 1. Then rn gets really small as n goes to infinity, so the limit converges and it converges to $$\frac{1}{1-r}$$

Are you looking for a rigorous proof of this, or is a description of the behavior like this sufficient?

3. May 30, 2013

### Lebombo

I was looking for a proof of where $\frac{1-r^{N+1}}{1-r}$ comes from and what it means. I think I found a proof for it, but still trying to absorb it.

4. May 30, 2013

### HallsofIvy

change the numbering: Let j= n-1 and n= j+ 1 so $r^n= r^{j+1}$. Of course, as n goes to 0, so does j. And it doesn't matter what "name" you give the parameter so just call it "n" now instead of "j".

Take a= 1, of course.