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Main content
Current time:0:00Total duration:4:55
AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.8 (EK)

Video transcript

let's think about the infinite series so we're going to go from N equals 1 to infinity of 1 over 2 to the N plus N and what I want to do is see if we can prove whether this thing it converges or diverges it is you can imagine based on the context of where this video shows up on Khan Academy that maybe we will do it using the comparison test and at any point if you feel like you can kind of take this to the finish line feel free to pause the video and do so so in order to to kind of figure out well get a sense for this series right over here it never hurts to kind of expand it out a little bit so let's do that so this would be equal to when N equals 1 this is going to be 1 over 2 to the 1 plus 1 so it's going to be 1 over 2 plus 1 it's going to be 1 3 plus that's N equals 1 when N equals 2 is going to be 1 over 2 squared which is 4 plus 2 plus 1 over 6 plus let's see when we go to when we go to 3 n equals 1 N equals 2 N equals 3 it's going to be 1 over 2 to the third which is 8 plus 3 is 11 so 1 over 11 maybe we'll do one more term 2 to the fourth power is going to be 16 plus 4 is 20 plus 1 plus 1 over 20 and obviously we just keep going on and on and on so in order to use so it looks it feels like this thing could converge all of our terms are positive but they are going they're getting smaller and smaller quite fast and if we really look at the if we look at the the behavior of this this the terms as n gets larger and larger we see that the 2 to the N in the denominator will grow much much faster than the N will so this will start to behave this kind of behaves like 1 over 2 to the N which is a clue of something that we might be able to use for the comparison test so let me just write that down so we have 1 over so the infinite series from N equals 1 to infinity of 1 over 2 to the N and so when N equals 1 here this is going to be equal to - this is going to be equal to 1/2 when n is equal to 2 this is going to be equal to 1/4 when n is equal to 3 this is equal to 1/8 when I need is equal to 4 this is equal to 1/16 and go on and on and on and on and what's interesting about this is we recognize that this is a geometric series so let me be clear this thing right over here that is the same thing as the sum from N equals 1 to infinity of 1/2 to the N power just writing it in a different way and since the absolute value of 1/2 which is just 1/2 so because because the absolute value of 1/2 is less than 1 we know that this geometric series converges we know that it converges and actually we even have formulas for finding the exact sum of or did find this exhibit figure out what it converges to and so we know this thing converges and we see that actually these two series combined meet all the constraints we need for the comparison test so let's go back to what we wrote about the comparison test so the comparison test we have two series all of their terms are greater than or equal to 0 all of these terms are greater than or equal to 0 and then in and for the corresponding terms in one series all of them are going to be less than or equal to the corresponding terms in the next one and so if we look over here we could consider this one the magenta series this is kind of our infinite series of dealing with a sub N and that this right over here is well I already did it in blue this is kind of the blue series and notice all their terms are non-negative and the corresponding terms one half is greater than 1/3 1/4 is greater than 1/6 1/8 is greater than 1 11 1 over 2 to the N is always going to be greater than 1 over 2 to the n plus n for the ends that we care about here and since we know that this converges since we know that the larger one converges it's a geometric series where the common ratio the absolute value of the common ratio is less than one since we know that the larger series converges therefore the smaller series or the one that's where every corresponding term is less than the one in the in the blue one that one must also converge so by the comparison test so by the comparison test the series in question must also must also converge
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