Why can't we use the formula a(r^n -1)/ (r-1) instead of a(1- r^n)/ (1-r) to find sum to infinity since same

Question

Why can't we use the formula a(r^n -1)/ (r-1) instead of a(1- r^n)/ (1-r) to find sum to infinity since they are actually the same formula?
Are there any restrictions when using either of the equations?
please try to explain and examples are appreciated!

how do you use a(r^n -1)/ (r-1) to find sum to infinity?

Answer 1

We absolutely can use either formula to find the sum of an infinite series.

It's preferable to use the formula

S = a(1 - r^n)/(1 - r)

Since, in an infinite series, the restriction is that |r| < 1.

But both are equally valid.

Answer 2

Let's just expand it and see what happens. First of all, let's get rid of a, it's irrelevant. a = 1. Okay.

(r^n-1)/(r-1) = r^(n-1) + r^(n-2) + ... while

(1-r^n)/(1-r) = 1 + r + r^2 + r^3 +...

For an infinite series, n = infinity. So, let's see what happens with the first series. The first term is

r^(infinity -1)

Which means if 0 < r < 1, then this would be zero. if 1 < r, then this would be infinity. Not a very good start. You could say, "okay, the first term is 0, we knew that. Let's go on with the next term."

r*(infinity - 2)

Well, no change, really. It's still either zero or infinity. This isn't very helpful.
In fact, you'll have to move over an infinity of terms before you get to the point where you have some numbers not either zero or infinity. On the other hand, the first term of the other series is:

1

Good start. We have a 1. The next term is

r

Great! Any value of r produces a value of r! The next is

r^2

Getting somewhere. We can actually start adding up some numbers. But there is just one problem. The higher n is, the larger

r^n

will be, tending towards infinity as n tends to infinity. For this reason, a series like this works best if -1 < r < 1. Then you have a bunch of manageable numbers that keep getting smaller with higher n, until somewhere you decide that you have an accurate enough answer, and disregard all the rest of the terms, an infinite of them. Now you see why mathematicians prefer using (1-r^n)/(1-r)?