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Infinite geometric series formula justification

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.