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Direct comparison test

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.8 (EK)
If every term in one series is less than the corresponding term in some convergent series, it must converge as well. This notion is at the basis of the direct convergence test. Learn more about it here.

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  • leafers ultimate style avatar for user Nahin Khan
    I don't understand why the sequence can't oscillate just because its terms are all non-negative.
    Can someone please explain why? I mean, for example, can't a sequence be like 3,4,3,4,3,4,3,4,3,4,3,4,... which is oscillating between two values but has all non-negative values??
    (35 votes)
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    • spunky sam blue style avatar for user Sal Khan
      From the author:In order for the sequence of partial sums to oscillate, you would need negative terms in the series. For example, to go from 4 to 3, you'd have to add -1. We made the assumption that every term in the sum is non-negative.
      (40 votes)
  • ohnoes default style avatar for user Cyan Wind
    , I don't quite understand how we can define that Σ a_n is "smaller" than Σ b_n? We compare the summations between them or what?
    (9 votes)
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    • blobby green style avatar for user Creeksider
      One of the assumptions established at the beginning is that each term in the "a" sequence is less than or equal to the corresponding term in the "b" sequence (that is, a_n is less than or equal to b_n). As a result, the sum of the terms in the "a" sequence has to be less than or equal to the sum of the terms in the "b" sequence.
      (8 votes)
  • blobby green style avatar for user 6phsi191
    Hello there,
    If one series is convergent, the another is also convergent.
    If one series is divergent, the other one is also divergent.
    or
    If one greater series is convergent, the another is also convergent.
    If one smaller series is divergent, the other one is also divergent.

    Which one do you want to mean in the video? I don't need to care about which is bigger or smaller, do I?

    Thanks
    Aung
    (4 votes)
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    • blobby green style avatar for user Steven
      The second version you said is correct.

      We say a series diverges if it adds all the way to infinity, right? It makes sense that if there's a series that diverges, a series larger than that one will also diverge. This happens because the smaller one went to infinity, so of course the bigger one will too. Similar thinking goes for convergent series. Convergence is defined as a series that adds up to a finite number. So if there's a series smaller than one that converges to a finite number, it wouldn't make sense for the smaller one to all the way up to infinity because a series larger than it converged.

      Hope that helped.
      (3 votes)
  • spunky sam green style avatar for user abdulaziz asiri
    what if b(n) diverges , will that make a(n) diverges.
    thanks
    (3 votes)
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    • leaf blue style avatar for user Stefen
      Since b(n) is ≥ a(n) for all n, if b(n) diverges, it says nothing about a(n) and the divergence test is not applicable.

      This is what you must understand about the divergence test.....
      If you have two different series, and one is ALWAYS smaller than the other, THEN
      1) IF the smaller series diverges, THEN the larger series MUST ALSO diverge.
      2) IF the larger series converges, THEN the smaller series MUST ALSO converge.

      You should rewatch the video and spend some time thinking why this MUST be so. Understanding this is paramount to moving forward with respect to understanding the divergence and convergence of series.

      Keep Studying and Keep Asking Questions!
      (4 votes)
  • leaf red style avatar for user Gunnar Grung Grotmol
    why does a_n,b_n have to be >= 0? Why can not both be nagative, just not equal to 0?
    (4 votes)
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  • aqualine seed style avatar for user ctymoszek
    At , Sal writes a[n],b[n] <=0.
    Is this just a given part of the problem i.e. is he saying "Imagine two series, a[n] and b[n] where both of them are >=0..."
    OR, did Sal use some information about the series to determine that they are nonnegative?
    (2 votes)
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    • leaf blue style avatar for user Stefen
      Sal is defining the properties of a[n] and b[n] so that the comparison test is valid and so that you can see the what is required of the series you wish to test in order to apply the "Comparison Test" - so your first thought is correct although this isn't a problem, it is (the basis of) a theorem.
      (3 votes)
  • female robot ada style avatar for user reshhhhwww233
    We always skip the case when the sequence contains negative numbers.
    So what test (or method) can be applied when we examine a sequence with negative numbers??
    for example,
    a sequence like,
    1, -0.1, 0.01, -0.001, 0.0001.........
    its series converge to 0.9090909090..................a repeating decimal.

    do we have a rule for them?
    (1 vote)
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  • mr pants teal style avatar for user Sophia Longo
    What's the intuition behind why if the smaller series diverges then the larger series diverges? Why not if the larger series diverges then the smaller series diverges too? And I have the same questions except regarding convergence instead of divergence.
    (1 vote)
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    • mr pink red style avatar for user andrewp18
      If 𝑥 > 𝑦 and 𝑦 > 𝑧, it follows that 𝑥 > 𝑧. So if we consider divergence to be approaching ∞, then if a series "diverges", and the partial sums of another series are larger than the original divergent series, then it follows that larger series must also be divergent.

      Using our previous inequalities, if we are given that the infinite series 𝑦 + 𝑦 + 𝑦 + ... diverges, then 𝑥 + 𝑥 + 𝑥 + ... also diverges since it is "larger than something that is already shown to be infinite" loosely speaking.

      Comment if you have questions!
      (2 votes)
  • piceratops seed style avatar for user MSabry490
    so does both laws can go viceverse?
    i mean if the greater one diverges does it mean that the smaller one also diverges or if the smaller one converges does it mean that the greater one converges?
    and if not so why did the test go that way?
    (1 vote)
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    • leafers sapling style avatar for user uyavuz96
      nope, think of this like 2 ordinary cups, one can fit inside of the other and this is like saying the inner cup can contain 100ml of water and the outer cup can contain only 50ml. it is impossible but the reverse is perfectly normal.
      (1 vote)
  • male robot hal style avatar for user Junior Bakshi
    If Σ a_n converges does Σ b_n converge ?
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

- [Voiceover] 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 one to infinity of a sub n. We're speaking in generalities here, and let's have another one. That's the series b sub n from n equals one to infinity, and we know some things about these series. The first thing we know is that all the terms in these series are non-negative. So a sub n and b sub n are greater than or equal to zero, which tells us that these are either going to diverge to positive infinity or they're going to converge 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. Once again, this is true for all the ns that we care about. So n equals one, two, three, all the way on, and on, and on. So the comparison test tells us that because all the corresponding terms of this series are less than the corresponding terms here, but they're greater than zero, that if this series converges, the one that's larger, if this one converges, well then the one that is smaller than it, or I guess when we think about it is kind of bounded by this one, must also converge. 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, I guess 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 sums to some finite value, then that tells us that the one that is in some ways I guess you could say smaller must also converge. So this one right over here must also converge. So that must also converge. So why is that useful? We'll see this in future videos. Well if you find if you're looking you have your a sub n and you're like 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 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. Of course it would only apply to the case where your original series, each of the terms are non-negative. Now what if you went the other way around? What if you could prove that the magenta series, the smaller one, and I guess I could kind of put them in quotes, this one right over here is 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 diverge because it oscillates between two values. Once again, if it's oscillating between values 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 our smaller series diverges, if this one diverges, then the larger one must also diverge. 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 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. We're going to start doing that in the next few videos.