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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.

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  • leaf red style avatar for user Paze
    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?
    (23 votes)
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    • spunky sam blue style avatar for user Ethan Dlugie
      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)
  • blobby green style avatar for user adamscarlat
    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)
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    • leaf blue style avatar for user Stefen
      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)
  • spunky sam blue style avatar for user Jai Sankar
    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?
    (5 votes)
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    • blobby green style avatar for user Creeksider
      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)
  • old spice man blue style avatar for user bb
    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)
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    • aqualine sapling style avatar for user Charles Russell
      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)
  • leafers tree style avatar for user Abdullah
    I think you forgot to put ar^(n+1) in the second series
    (4 votes)
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    • leaf blue style avatar for user Stefen
      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)
  • mr pink red style avatar for user Rainman
    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)
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  • piceratops seedling style avatar for user Lily G.
    So, this particular formula would work if "r" were anywhere between -1 and 0, and 0 and 1, right?
    (5 votes)
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  • blobby green style avatar for user Dominic Chan
    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)
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    • purple pi purple style avatar for user doctorfoxphd
      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)
  • leaf green style avatar for user Coco
    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)
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    • leaf blue style avatar for user Stefen
      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)
  • aqualine ultimate style avatar for user christophorino
    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)
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    • purple pi purple style avatar for user redthumb.liberty
      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.