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Main content
Current time:0:00Total duration:4:30
AP.CALC:
LIM‑7 (EU)
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LIM‑7.A (LO)
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LIM‑7.A.8 (EK)

Video transcript

so let's get a basic understanding of the comparison test when we are trying to decide whether a series is converging or diverging so let's think of two series so let's say that I have this magenta series here it's an infinite series from N equals 1 to infinity of a sub N or speaking in generalities here and let's say I have another one and that's the series B sub n from N equals 1 to infinity and we know some things about these series the first thing we know is that all of the terms in these series are non-negative so a sub N and V sub n are greater than or equal to 0 which tells us that this that these are these are going to diverge to positive infinity or they're going to converge to some to some finite value they're not going to oscillate because you're not going to have negative values here you can't go to negative infinity because you don't have negative values here now let's say we also know that each of the corresponding terms in the first series are less than or equal to the corresponding term in the second series less than or equal to B sub N and once again this is true for all the ends that we care about so N equals 1 2 3 all the way on and on and on so the comparison test tells us that because this series all the corresponding terms of this series are less than the corresponding terms here but they're greater than 0 that if this series converges the one that's larger if this one converges when the one that is smaller than it or guess 1 we think about is kind of bounded by this 1 must also converge and I'm not doing a formal proof here but hopefully that gives you a little bit of intuition so the comparison test tells us if if I guess what in my brain the larger series the one whose corresponding terms are at least as large as the ones here if this one converges if this one doesn't go unbounded towards infinity it goes to its sums to some finite value then that tells us that the 1 that is in some ways I guess you say smaller must also converge must also so this one this one right over here must also converge so that must also converge so why is that useful and we'll see this in future videos well if you find if you're looking say you you have your a sub n and you're like hey gee I wish I could prove that it converges I kind of have a gut feeling it converges the comparison test tells us well just find another series that is always that whose corresponding terms are at least as large as the corresponding terms here and if you can prove that one converges then you're good with this one and of course it would only apply to the case where your original series each of the terms each of the terms are non-negative now what if we want the other way around what if you could prove that the magenta series the smaller one and I guess I could I could kind of put them in quotes you know this one right over here is the smaller the smaller I guess each of its corresponding terms are smaller what if you could prove this one diverges well if this one diverges it's going to go unbounded to infinity it's not going to go to negative infinity all the terms are positive it's not going to oscillate it's not going to diverge because it oscillates between two values once again if it's oscillating between values that wouldn't the only way you could do that is if you had negative terms here so this would kind of be unbounded towards infinity well if this one is unbounded each of these corresponding terms are larger so this one must also be unbounded so let's write that down so the comparison test tells us if if our smaller smaller series diverges if this one diverges then the larger one must also diverge then the month larger one must also diverge and so once again if you wanted to prove that this thing right over here is going to diverge and if you have a once again and you know that all the B sub NS are greater than or equal to zero and you want to prove it diverges well maybe you could try to find another series where each of the corresponding terms are less than the corresponding terms here and you could prove this one diverges then you would be all set and we're going to start doing that in the few videos