AP®︎/College Calculus BC
In some cases where the direct comparison test is inconclusive, we can use the limit comparison test. Learn more about it here.
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- Did Sal ever explain the intuition behind the limit comparison test?(23 votes)
- It's actually straightforward if you think about the definition of a limit. If you have lim n->inf a/b = L then L - epsilon < a/b < L + epsilon. Apply algebraic manipulation and we get that either b*(L-epsilon) < a or a < b*(L+epsilon). Since L+epsilon is just a constant, it won't change the convergence of b. And now we can use the regular comparison test. That means that in the first case ( b(L*eps) < a), if b diverges, then a diverges, and in the second case ( a < b(L*eps)), if b converges, a converges.(25 votes)
- I didn't fully get how to pick the series Bn.
I understand that it's convenient for it to be a series that we know for sure if it diverges or converges, but surely we can't pick one randomly.
What proprieties do we want Bn to have?(16 votes)
- I think it sometimes helps to find the dominant term in the numerator and/or denominator. It is best not to choose a Bn at random because that Bn might not be that similar to An. if you have An as (5n^2) / (n^3 -n^2-n) then you should choose Bn as (5n^2) / (n^3) and use the limit comparison test since An > Bn and you are not sure if An diverges to use the regular comparison test. Bn converges by the rules of a p-series but that does not tell you anything about An... I think (:(3 votes)
- What dose it mean if the limit of a sub n divided by b sub n as n approaches infinity equals zero? is the test inconclusive?
*lim(n goes to infinity)(a (sub n)/b (sub n))=0: inconclusive?*(6 votes)
- Actually, there are two other conclusion that can be made from this test. If lim n->infinity for a sub n divided by b sub n=0, and you know that b sub n converges then a sub n also converges. If b sub n diverges then the test is inconclusive. Also you can find that lim n->infinity a sub n divided by b sub n= infinity and b sub n diverges then a sub n diverges.(6 votes)
- How do I know the order of the division, is it always An/Bn (so unknown/known) or bigger/smaller?(6 votes)
- I have the same question. I know this comment is about 6 years old but did you ever end up figuring it out? Thanks!(4 votes)
- What's the difference between the limit comparison test and the comparison test? I mean in both aren't you comparing the rate of each Series?(5 votes)
- In the comparison test you are comparing two series Σ a(subscript n) and Σ b(subscript n) with a and b greater than or equal to zero for every n (the variable), and where b is bigger than a for all n. Then if Σ b is convergent, so is Σ a. If Σ a is divergent, then so is Σ b.
In the limit comparison test, you compare two series Σ a (subscript n) and Σ b (subscript n) with a n greater than or equal to 0, and with b n greater than 0. Then c=lim (n goes to infinity) a n/b n . If c is positive and is finite, then either both series converge or both series diverge.
In other words, in the limit comparison test you do not know whether your series converge/diverge, so using limits you find whether they both will diverge or converge. In the comparison test, you know whether on converges/diverges and using that knowledge, attempt to find whether the other converges or diverges.
Hope this helped. You can read more here: http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx(5 votes)
- Why does the limit have to be positive and why do both functions have to be greater than zero over the interval? I understand if you had a positive and a negative function that both approached zero, you would be looking at graphs that were (sort of) reflections across the x axis, but why would this invalidate the limit comparison test?(5 votes)
- At6:40, Sal says that 1/2^n converges because it is a geometric series with a common ratio of less than 1. But there are other series which also have these properties and are not convergest, like 1/n.(1 vote)
- At4:00, is zero consider a positive integer ? If [an/bn = 0] can we conclude on the convergence of the series ?(3 votes)
- No, zero is not considered a positive integer. In fact, some mathematicians consider zero to not even be finite. As for the answer to your question, the test would be inconclusive.(1 vote)
- ins't 1/2^n -1 larger than 1/2^n ? I thought you needed to compare the original to a larger series in order to confirm that is diverges or converges? or is that different on the limit test?(2 votes)
- If the larger thing converges, then the smaller thing must converge.
If the smaller thing diverges then the larger thing must diverge.
You must determine the complete and proper context for the comparison.(2 votes)
- Let's remind ourselves, give ourselves a review of the Comparison Test, see where it can be useful, and maybe see where it might not be so useful, but luckily we'll also see the Limit Comparison Test, which can be applicable in a broader category of situations. So we've already seen this, we want to prove that the infinite series from n equals one to infinity of one over two to the n plus one converges how can we do that? Well, each of these terms are greater than or equal to zero, and we can construct another series where each of the corresponding terms are greater than each of these corresponding terms. And that other series, the one that jumps out at, that will likely jump out for most folks would be one over two to the n, one over two to the n is greater than, is greater than, and all we would have to really say is greater than or equal to, but we could actually explicitly say it's, oh I'll just write it's greater than or equal to, one over two to the n plus one for, n is equal to one, two, all the way to infinity - why? Because this denominator is always going to be greater by one if you're denominator is greater, the overall expression is going to be less, and because of that, because each of these terms are, they're all positive, this one, each corresponding term is greater than that one, and by the Comparison Test, because this one converges, this kinda provides an upper bound, because this series we already know converges, we can say, so because this one converges, we can say that this one converges. Now, let's see if we can apply a similar logic to a slightly different series. Let's say we have the series the sum from n equals one to infinity of one over two to the n minus one. In this situation can we do just the straight up Comparison Test? Well no, because you cannot say that one over two to the n is greater than or equal to one over two to the n minus one. Here, the denominator is lower, means the expression is greater, means that this can't, each of these terms can't provide an upper bound on this one. This one is a little bit larger, but on the other hand you're like, 'Ok, I get that.' But look, as n gets large, the two to the n is going to really dominate the minus one or the plus one, or this one has nothing there, this one just has two to the n. The two to the n is really going to describe the behavior of what this thing does. And I would agree with you, but we just haven't proven that it actually works, and that's where the Limit Comparison Test comes in helpful. So let me write that down, 'Limit, Limit Comparison Test, 'Limit Comparison Test', and I'll write it down a little bit formally, but then we'll apply it to this infinite series right over here. So what Limit Comparison Test tells us, that if I have two infinite series, so this is going from n equals k to infinity, of a sub n, I'm not going to prove it here, we'll just learn to apply it first. This one goes from n equals k to infinity, of b sub n. And we know that each of the terms, a sub n, are greater than or equal to zero, and we know each of the terms, b sub n, are actually we're just going to say, greater than zero, because it's actually going to show up in the denominator of an expression, so we don't want it to be equal to zero, for all the n's that we care about. So for all n equal to k, k plus one, k plus two, on and on, and on and on, and, and this is the key, this is where the limit of the Limit Comparison Test comes into play, and, if the limit, the limit as n approaches infinity, of a sub n over b sub n, b sub n is positive and finite, is positive and finite, that either both series converge, or both series diverge, so let me write that. So then, that tells us that either, either both series converge, or both diverge, which is really really useful. It's kind of a more formal way of saying that, 'Hey look, as n approaches infinity, if these have similar 'behaviors then they are either going to 'converge, or they're both going to diverge.' Let's apply that right over here. Well if we say that our b sub n is one over two to the n, just like we did up there, one over two to the n, so we're going to compare, so these two series right over here, notice it satisfies all of these constraints, so let's take the limit, the limit as n approaches infinity of a sub over b sub n, so it's going to be one over two to the n minus one over one over two to the n, and what's that going to be equal to? Well that's going to be equal to the limit as n approaches infinity, of two, if you divide by one over two to the n that's just going to be the same thing as multiplying by two to the n, so it's going to be two to the n over two to the n minus one. Over two to the n minus one, and this clearly, what's happening in the numerator and the denominator, these are approaching the same, well we can even write it like, we can even write it like this, divide the numerator and denomiator by two to the n, if you want, although it's probably going to jump out at you at this point. So, limit as n approaches infinity, let me just scroll over to the right a little bit. If I divide the numerator by two to the n, I'm just going to have one. If I divide the denominator by two to the n, I'm going to have one minus one over two to the n. And now it becomes clear, this thing right over here is just going to go to zero, and you're going to have one over one. The important thing is that this limit is positive and finite because this thing is so this thing right over here, is positive and finite, the limit is one is positive and finite, if this thing converges and this thing converges, or this thing diverges, and this diverges, well, we already know this thing converges, it's a geometric series where the common ratio is less than one, and so therefore this must converge as well, so that converges as well.