# Thread: Geometric Mean, infinite sum when -1 < r < 1

1. ## Geometric Mean, infinite sum when -1 < r < 1

I'd like to know the reason behind why there's an infinite term sum behind a geometric sequence when -1 < r < 1

2. Originally Posted by Masterthief1324
I'd like to know the reason behind why there's an infinite term sum behind a geometric sequence when -1 < r < 1
Here's the standard proof: first look at finite sums because $\displaystyle \sum_{i=1}^\infty a_i$ is defined as the limit, as n goes to infinity of $\displaystyle \sum_{i=1}^n a_i$.

$\displaystyle S= \sum_{i=0}^n ar^i= a+ ar+ ar^2+ ar^3+ \cdot\cdot\cdot+ ar^n$. Subtract a from both sides: $\displaystyle S- a= ar+ ar^2+ ar^3+ \cdot\cdot\cdot+ ar^n$.

Factor an "r" out of that sum: $\displaystyle S- a= r(a+ ar+ ar^2+ \cdot\cdot\cdot+ ar^{n-1})$. Divide both sides by r: $\displaystyle \frac{S- a}{r}= a+ ar+ ar^2+ \cdot\cdot\cdot+ ar^{n-1}$.

That is almost what we started with- it is missing "$\displaystyle ar^n$" but we can add that to both sides:
$\displaystyle \frac{S-a}{r}+ ar^n= a+ ar+ ar^2+ \cdot\cdot\cdot+ ar^n= S$.

Because these are finite sums, all those are legal algebraic operations and we have now arrived at $\displaystyle \frac{S-a}{r}+ ar^n= S$. Now solve for S.

$\displaystyle \frac{S-a}{r}= S- ar^n$
$\displaystyle S-a= rS- ar^{n+1}$
$\displaystyle S- rS= a- ar^{n+1}$
$\displaystyle (1- r)S= a(1- r^{n+1}$
$\displaystyle S= \frac{a(1- r^{n+1})}{1- r}$

That is the sum of a finite number, n, of terms. The only place n occurs in that formula is as "$\displaystyle r^{n+1}$". If r> 1 that will get larger and larger without bound as n increases so we cannot "take the limit" as n increases. If r< -1, then we have alternating positive and negative values but still getting larger and larger without bound in absolute value so we cannot "take the limit" as n increases.

If -1< r< 1, then r^n gets smaller and smaller- its limit is 0 and so the limit of S is $\displaystyle S= \frac{a(1- 0)}{1- r}= \frac{a}{1- r}$.

Finally, if r= 1 we just have S= a+ a+ a+ ....+ a = na which gets larger and larger without bound (for all a except 0) and so has no limit. If r= -1, we get S= a- a+ a- a+/cdot/cdot/cdot$\displaystyle \pm$a which is a if n is odd and 0 if n is even- still no limit.

That limit exist only if -1< r< 1, in which case the sum is $\displaystyle \frac{a}{1-r}$ or if a= 0 in which case the sum is 0 no matter what r is.

3. Thanks for the reply, I'm in the process of understanding your explanation though I need some things clarified.

I've never used the 'E' notation before what does it mean or how is it used in mathematics? (It's not important but I'd like to know).

I'm stuck at:
$\displaystyle S- rS= a- ar^{n+1}$
What property did you use to simplify it to:

Just to clarify -- in the simplest of terms, when you add the numbers of the sequence, in order, infinitely, there is a finite sum?
It just occurs to me as odd because the statement seems to contradict.

4. Originally Posted by Masterthief1324
Thanks for the reply, I'm in the process of understanding your explanation though I need some things clarified.

I've never used the 'E' notation before what does it mean or how is it used in mathematics? (It's not important but I'd like to know).

I'm stuck at:
$\displaystyle S- rS= a- ar^{n+1}$
What property did you use to simplify it to:
Factorisation. The terms of the left hand side have a common factor S. The terms of the the right hand side have a common factor of a.

5. $\displaystyle \sum\ is\ a\ Greek\ symbol\ meaning\ 'sum'.$