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what I want to do is another proof II like thing to think about the sum of an infinite geometric series and it will use a very similar idea to what we use 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 zero and we're going to go we're never going to stop we're going to go 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 K power to the K power actually we do K in that color K equals zero all the way to infinity and so let's just call this let's just call this thing right over here let's call this s sub let's call this s sub infinity we're going all the way to infinity right over here and we're going to and so this if we were to expand it out is going to be equal to it's going to be equal to a times R to the 0 I'll let me just write it out like that which is just just a a times R to the 0 power plus a times R to the first power R to the first power plus a times R to the second power R to the second power plus and we could just keep going let me do one more plus a actually we could just keep going on and on and on I think you get the general idea now just like when we try to find drive a formula for the sum of a finite geometric series we just said well what happens if you take this 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 there's what I said this is proof e is this is not always clear what it's it's a little bit when you're multiplying something times infinite terms or infinite sum at least this will at least give you the general idea when you start think about infinity sometimes I have to think about things a little bit deeper so R time's this infinite sum well that's going to be equal to we're going to multiply every term here times R so a R to the 0th power times R is going to be a times R a times R to the first power multiply this 1 times R you're going to get a times R to the second power a times R to the 2nd power think you see where this is going multiply this 1 times R you're going to get plus a times R to the 3rd power and we would just keep on going we just keep on going so let me just show that so plus 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 we could express that as our sum S sub infinity minus our common ratio times s sub infinity s sub infinity is going to be equal to so when you subtract you're going to have a times R to the 0 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 a times 1 which is a let me write that same color equal to a but every other term you're going to have a times R to the first but then you're subtract a times R to the first you're going to have a times R to the second we're going to subtract a times on a second so every other term is going to be subtracted is going to be subtracted away this happens all the way 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 there are some 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 a 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 we are going to get a finite value so this is going to be equal to a this is going to be equal to a over over 1 minus R so once again it's kind of neat if I was to say I had let's say I had the sum and I don't say we started with 5 and then each time we were to multiply by I don't know 3/5 so 3 so 5 plus 3/5 times 5 is 3 times 3/5 is going to be 9 fifths 9 fifths or I multiply by 3/5 again then I'm sorry not 9 fifths so 5 my brain isn't working right 5 times 3/5 is going to be 3 times 3/5 is going to be is going to be 3 times this it's going to be 9 fifths actually now 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 this is going to be 81 over 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 or 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 this is going to be equal to 5 over 2/5 which is the same thing as 5 times times 5 over 2 which is 25 over 2 which is equal to 12 and 1/2 or 12.5 once again amazing result I'm taking an infinite 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 in this did it even works out a math matically in this denominator that you are going to get a 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 really dealing we're not anymore dealing strictly with a geometric series anymore but obviously if R was 0 then everything 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 time you just said it would be a then clearly you just be left with a is the sum and a over 1 minus 0 is still a so this 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 a minus a plus a minus a and so the sums 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 you're your common ratio is between 1 and negative 1

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