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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 7

Lesson 6: Infinite geometric series- Worked example: convergent geometric series
- Worked example: divergent geometric series
- Infinite geometric series
- Infinite geometric series formula intuition
- Proof of infinite geometric series as a limit
- Infinite geometric series word problem: bouncing ball
- Infinite geometric series word problem: repeating decimal
- Convergent & divergent geometric series (with manipulation)

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# Infinite geometric series formula intuition

Sal uses a clever algebraic manipulation to find an expression for the sum of an infinite geometric series. Created by Sal Khan.

## Want to join the conversation?

- There's something wrong with my calculations; somebody please help me.

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?(24 votes)- 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.(33 votes)

- 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!(17 votes)- 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!(15 votes)

- At7:27, 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?(5 votes)
- 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.(14 votes)

- 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(6 votes)- 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.(6 votes)

- I think you forgot to put ar^(n+1) in the second series(4 votes)
- 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:

https://www.youtube.com/watch?v=elvOZm0d4H0(8 votes)

- 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?(2 votes)
- Up to the 8th term would imply that you are only summing the first 8 terms, not the entire series.(3 votes)

- So, this particular formula would work if "r" were anywhere between -1 and 0, and 0 and 1, right?(5 votes)
- 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.(3 votes)

- 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)(2 votes)- 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)`

(7 votes)

- 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?(2 votes)
- 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

https://www.youtube.com/watch?v=elvOZm0d4H0 - this is about the various types of infinities

Good luck and keep studying!(6 votes)

- 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(2 votes)- 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.(5 votes)

## Video transcript

What I want to do is
another "proofy-like" thing to think about the sum of
an infinite geometric series. And we'll use a
very similar idea to what we used to find the sum
of a finite geometric series. So let's say I have a geometric
series, an infinite geometric series. So we're going to
start at k equals 0, and we're never going to stop. We're it's going all
the way to infinity. So we're never going to
stop adding terms here. And it's going to be our first
term times our common ratio. Our common ratio
to the kth power. Actually let me do
k and that color. k equals 0 all the
way to infinity. And so let's just call
this thing right over here, let's call this s sub infinity. We're going all the way to
infinity right over here. And so this, if we
were to expand it out is going to be equal to a
times r to the 0-- actually let me just write it out
like that which is just a. a times r to the 0 power plus
a times r to the 1st power. r to the 1st power. Plus a times r to the 2nd power. r to the 2nd power. Plus-- and we could just
keep going on and on and on. I think you get
the general idea. Now just like when
we tried to derive a formula for the sum of
a finite geometric series we just said, well what
happens if you take the sum and if you were to
multiply every term by your common ratio. Every term by r. So let's do that. Let's imagine this sum. And we're going to
multiply every term by r. And the reason
why I said this is "proofy" is this is not always
clear-- It's a little bit, when you're multiplying
something times infinite terms or an infinite sum, at
least this will be at least give you the general idea. Or when you start thinking
about and infinity, sometimes I just think about
things a little bit deeper. So r times this infinite sum? Well that's going to
be equal to-- We're just going to multiply
every term here times r. So a r to the 0'th power times
r is going to be a times r. a times r to the 1st power. Multiply this one times r? You're going to get a
times r to the 2nd power. a times r to the 2nd power. I think you see
where this is going. Multiply this one times r? You're going to get plus a
times r to the 3rd power. And we would just keep on going. We'd just keep on going. So let me just show that. So plus dot dot dot. Now what happens if
we were to subtract this sum from this top sum? So on the left
hand side, we could express that as our sum s sub
infinity minus our common ratio times s sub infinity. Is going to be equal to-- So when you subtract you're
going to have a times r to the 0'th power, which is
really just the same thing as a. That's just going to be
a. a times r to the 0 is just a times 1 which is a. We'll write in that same color. Is equal to a. But every other
term, you're going to have a times r to
the 1st, but you're gonna subtract a
times r to the 1st. You have a times r to
the 2nd, but you're going to subtract a
times r to the 2nd. So every other term is
going to be subtracted away. And this happens all
the way to infinity. It never, never ends. So the only term
that you're left with is just that
first one, is just a. And so now we can actually
try to solve for our sum. If you factor out
the s sub infinity, you are left with 1 minus r. 1 minus r. s times s, our sum, times
1 minus r is equal to a. Divide both sides
by 1 minus r, and we get that our sum, the
thing that we cared about-- And once again, this is
kind of an amazing result. That we're taking the sum of
an infinite number of terms and under the
proper constraints, we are going to
get a finite value. So this is going to be
equal to a over 1 minus r. So once again,
it's kind of neat. If let's say I had the sum,
let's say we started with 5, and then each time we
were to multiply by 3/5. So 5 plus 3/5 times 5 is 3,
times 3/5 is going to be 9/5. 9/5, or I'll multiply by 9/5
again-- Oh, sorry not 9/5. My brain isn't working right! 5 times 3/5 is going
to be 3 times 3/5. Is going to be-- 3
times this is going to be 9/5-- actually
that was right. My brain is working right. Times 3/5 is going
to be 27 over 25. Times 3/5 is going to be 81/125. And we keep on going on
and on and on forever. And notice these
terms are starting to get smaller and
smaller and smaller. Well actually all
of them are getting smaller and smaller and smaller. We're multiplying
by 3/5 every time. We now know what the
sum is going to be. It's going to be our
first term-- it's going to be 5-- over 1
minus our common ratio. And our common ratio
in this case is 3/5. So this is going to be
equal to 5 over 2/5, which is the same
thing as 5 times 5/2 which is 25/2 which is equal
to 12 and 1/2, or 12.5. Once again, amazing result. I'm taking a sum of
infinite terms here, and I was able to
get a finite result. And once again, when
does this happen? Well, if our common ratio--
if the absolute value of our common ratio--
is less than 1, then these terms
are going to get smaller and smaller and smaller. And you'll even see here it
even works out mathematically in this denominator
that you are going to get a reasonable answer. And it makes sense
because these terms are getting smaller
and smaller and smaller that this thing will converge. Even if r is 0. If r is 0, we're still
not dealing strictly with a geometric series anymore,
but obviously if r was 0, then you're really only going
to have this-- well, even this first term is
kind of under debate depending on how you
define what 0 to 0 is. But if your first term
you just said would be a, then clearly you'd just be left
with a is the sum, and a over 1 minus 0 is still a. So this formula that we just
derived does hold up for that. It does start to break down if
r is equal to 1 or negative 1. If r is equal to 1 then
as you imagine here, you just have a plus
a plus a plus a, going on and on forever. If r is equal to negative 1
you just keep oscillating. a, minus a, plus a, minus a. And so the sum's value keeps
oscillating between two values. So in general this
infinite geometric series is going to converge
if the absolute value of your common ratio
is less than 1. Or another way of saying
that, if your common ratio is between 1 and negative 1.