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### Course: Algebra (all content)>Unit 18

Lesson 7: Infinite geometric series

# Infinite geometric series formula intuition

In the video, we learn about the sum of an infinite geometric series. The sum converges (has a finite value) when the common ratio (r) is between -1 and 1. The formula for the sum is S = a / (1 - r), where a is the first term. Created by Sal Khan.

## Want to join the conversation?

If we take the ratio to be 2, then the result of the sum would be +infinite.
But let's put it in numbers in the same way Sal did:

X = 5 + 5*2 + 5*2² + 5* 2³ etc....
now we multiply X by r, which is 2, and then let's subtract them.
Now, X-2X = 5
X=5/1-2
X=-5 (!)

What's wrong with this logic?
It should be +infinite, right?
• Sal said it himself at the end, that the common ratio _r_ must satisfy the condition |r|<1. This is not the case for your specific sum. Dealing with infinity is in general a dangerous venture and can get you into a lot of trouble if you don't treat it vigorously.
Here is a simple yet interesting example I found on wikipedia:
∑ 0 from n=1...oo (oo denotes infinity)
This sum is clearly 0, but we can do a little math trickery...
=∑ (1-1) from n=1...oo
=(1-1)+(1-1)+...=1+(-1+1)+(-1+1)+...=
1+∑(-1+1) from n=1...oo
=1+∑0
=1
Which is definitely not right.
• In the derivation of the finite geometric series formula we took into account the last term when we subtracted Sn-rSn and were left with a-ar^(n+1) in the numerator. Here Sal subtracted Sinf-rSinf and sort of ignored the last term and just had the numerator to equal a.
My question is, in the case of the infinite series, how can you rigorously prove that every term does cancel out?
Thanks!
• Ooooh - the mysteries of infinity! The thing of it is, THERE IS NO LAST TERM. That means, with respect to Sn and rSn, all terms cancel out except the first term in Sn, a_0 since for all other terms Sn has just as many as rSn.
Part of the formality you may be missing is noting that as n goes to infinity, the limit of each term a_n which is composed of (a_0)r^n goes to zero. Here is another way of writing the proof that uses the limit argument to get to the same place Sal did: http://bajasound.com/khan/khan0009.jpg.
To understand the proof, you need to understand that for |r|<1 the limit of r^n as n-->infinity is zero.
Great Question!
Keep Studying!
• At , Sal said that if r= -1, then the values would keep on oscillating. However, if you actually work it out, the series would converge to a/2. If we say that the series, is S, than S= a-a+a-a+a-a+a-a+a... . Also, a-S would equal a-a+a-a+a-a+a... . This is equal to the original series S. So, we can say a-S=S. Then we get a= 2S. Therefore, S=a/2. Is there something I am doing I am not disregarding in my calculuations?
• Your logic seems plausible but fails the epsilon-delta test, which is the ultimate test for whether a series converges. There is no delta for which larger values of n will produce S values closer than, say, a/4 (a possible epsilon). We know this because it's clear that there is no point beyond which the sum stops oscillating between a and 0. Your result of a/2 is the average of the two values between which the sum oscillates, not the value to which the sum converges.
• Please, help me understand the parts in this proof:
why do we multiply and after that, we subtract the first sum from the second sum, because these two steps give us the formula, but based on what reasoning we do this exact step: first multiply by the common ratio and second - subtract the second sum from the first sum?
thanks
• If you divided a by r the first term would become a/r which is no good. By multiplying both sides by r you create a kind of phase shift of terms such that by subtracting like terms from both series you are left with a first term and a last term that goes on to give the required sum.
• I think you forgot to put ar^(n+1) in the second series
• We don't need to since both series have an ar^(n+1) term, which cancel out, leaving only the ar^0 term. If that seems weird to you, and if this is your first exposure to sequences/series that go to infinity, then the mathematical concept of infinity can seem a bit strange.
Try this video to get another perspective on the concept of infinity:
• what will be the sum of infinite geometric series 2/3 + 1/3 + 1/6..... up to 8 term? and what does that mean up to 8 term?
• Up to the 8th term would imply that you are only summing the first 8 terms, not the entire series.
• So, this particular formula would work if "r" were anywhere between -1 and 0, and 0 and 1, right?
• That is correct. If the absolute value of r is 1 or greater, that is if r IS NOT -1<r<0 or 0<r<1, (as you observe) then the sum diverges (goes to infinity) and the formula will not give a correct answer.
• what if u times r^2 to S(n)

then won't you get S(infinity) - r^2 S(infinity) = to ar^(0) + ar^(1)
and S(infinity) = (ar^(0) + ar^ ( 1 ) ) / (1 - r^2)
• Well, let's try that:
Starting with
S∞ = ar⁰ + ar¹ + ar² + ar³ + ar⁴ + ar⁵ + ar⁶ + ar⁷ . . .
multiply that series by r²
r²∙S∞ = r²∙ar⁰ + r²∙ar¹ + r²∙ar² + r²∙ar³ + r²∙ar⁴ + r²∙ar⁵ + . . .
Simplify
r²S∞ = ar² + ar³ + ar⁴ + ar⁵ + ar⁶ + ar⁷ + . . .
Then subtract
S∞ - r²S∞
S∞ = ar⁰ + ar¹ `+ ar² + ar³ + ar⁴ + ar⁵ + ar⁶ + ar⁷`
- r²S∞ = `- ar² - ar³ - ar⁴ - ar⁵ - ar⁶ - ar⁷ `
Resulting in
S∞ - r²S∞ = ar⁰ + ar¹ which is equivalent to a + ar
S∞(1 - r²) = a + ar
and
S∞ = (a + ar)/(1 - r²) `your result`
That isn't simplified, though:
You can factor further, I guess
S∞ = a(1 + r)/[(1 + r)(1 - r)]
Cancel the like factors of (1 + r)
Back to
`S∞ = a/(1 - r)`
• In order to get all terms to cancel except for a, we must assume that Sinfinity has one extra term than r*S infinity. Someone said there is no last term in an infinite series. But I am not talking about a last term, just about the fact that Sinfinity must have one extra term (if there isn't the cancellation does not work). Since r*Sinfinity has one less term, it is finite (or has an upper bound). So is r, if <1, an operator on infinite series, bounding them, or making them finite?
• Infinity is not a number, it is a concept. S and rS have the same number of terms. The difference is that S has a term that rS does not. What happens is when you get way way way out there, that is, the value of n for S_n and rS_n is very very very large, the value of each term of S_n and rS_n become negligible. If you do not think each term is small enough to be accurate for your needs, then keep on increasing the size of n until it does satisfy your needs.

The next weird thing is that if you let n go to infinity, the sequences n and 2n have the same number of terms!

I highly recommend that you spend some time in personal research on the concept of infinity.
Here are some links to get you started:
http://www.emis.de/proceedings/PME30/4/345.pdf - this is about typical ways infinity is understood by students

Good luck and keep studying!
• At the end, Sal said that 5/2/5 is 12.5. But isn't it one half? 5/2*5, 5/10, 1/2
Also 5/5 is 1, 1 halved is 1/2
• You need to be very careful with multiple divisions or divisions of fractions.

`5/2/5` is actually `5/1 ÷ 2/5`, which is evaluated:

``5/1 * 5/2 = 25/2 = 12.5``

I always make it a point to model those types of fraction divisions as:

``5/(2/5) → 5*(5/2)``

That helps to keep me from falling into the invalid reduction trap that you just fell into.