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Comparison tests

# Limit comparison test

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.9 (EK)
In some cases where the direct comparison test is inconclusive, we can use the limit comparison test. Learn more about it here.

## Want to join the conversation?

• Did Sal ever explain the intuition behind the limit comparison test?
• 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.
• 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?
• 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 (:
• 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?*
• 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.
• 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?
• 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
• How do I know the order of the division, is it always An/Bn (so unknown/known) or bigger/smaller?
• I have the same question. I know this comment is about 6 years old but did you ever end up figuring it out? Thanks!
• 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?
• At , 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)
• Observe that the harmonic series `∑ 1/n` is not geometric.
• At , is zero consider a positive integer ? If [an/bn = 0] can we conclude on the convergence of the series ?
• 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)
• what will be Nth term for the series : (1/3)+(1*2/3.5)+(1*2*3/3*5*7)+......
(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?