The Limit Comparison Test

This video explains how to apply the limit comparison test to determine if an infinite series converges or diverges.

[TrumpetHX]

Thought I was going to fail my calc 2 test tomorrow. You are a GOD.

[Mathispoweru]

Hello again, you are correct that the limit would be different but by the limit comparison test as long as the limit is positive and finite then both series converge or diverge. So with a limit of 1/4 or 4, both would converge or both would converge. Does that make sense? Some books only state it one way but either works. I'm happy to know you have found my video helpful. I appreciate you letting me know.

[eleazaralmazan]

You are the best math instructor on YouTube!

[Mathispoweru]

We can actually put the known convergent or divergent series in the numerator or denominator. Notice by putting 3/4^n in the numerator, it makes simplifying the ratio easier than if 3/(4^n-3) was in the numerator. However, it would probably work either way.

[Mathispoweru]

Because (3/4^n) is a geometric sequence with R = 1/4, which we know converses. We need to use a sequence we know converses to use the limit comparison test if we are using the limit comparison test to show convergence. In the second example, we are trying to show divergence so we use a known divergent series. I hope that helps.

[Mathispoweru]

@sakuranbo888 dummiesdotcom explains your question very well. "Contrary to the formal definition of the Limit Comparison Test, the limit, L, doesn’t have to be finite and positive for the test to work. If the benchmark series is convergent and the limit is zero, then your series must also converge. If the limit is infinity, you can’t conclude anything. If the benchmark series is divergent and the limit is infinity, then your series must also diverge. If the limit is zero, you learn nothing.

[Mathispoweru]

It depends on the series you are comparing to. If the limit is finite and positive, both are convergent or both are divergent. It depends if you are using the test with a known convergent or known divergent series.

[jasonjkeller]

@ 8:09 you changed the numerator n^4+2n^2-1 to n^4+2n^2+1, same thing at 9:27 when you rewrote the final answer. Once again though, an excellent and helpful video. Thanks!

[WickedSweet]

Thank you so much, you just solved all my problems with this chapter (just in time for the test tomorrow!)
Thanks again!!!

[dpmike]

at 6:47 you said it is finite and divergent ... shouldn't it be convergent?

[msos]

Thank you very much you saved my life :)

Best instructor ever!

[Mathispoweru]

@jasonjkeller79 Thank you again for noticing. Luckily it doesn't affect the limit. My writing is so small I can't annotate the change.

[preciousrosej.]

Logged in just to upvote! Great work!

[Mathispoweru]

Actually, it doesn't matter. Either they both converge or both diverge. You can switch An and Bn

[gder]

For example #2, 5/((n^2)-1)^.5, could we use the direct comparison test as well? comparing the series to a smaller series (1/n) that diverges.

[puravpatel]

YES!! It does matter, i was doing problem and i got the answer to be 4 and i checked the answer and it was 1/4. So i flipped in the start i would of got the answer. Otherwise i like watching your videos, they are almost better than Khan and PatrickJMT. You are good at explaining stuff and your examples are like the ones from the Calc book. KEEP UP THE GOOD WORK!

[VileVendetta]

Because it's arbitrary what you make a sub n or b sub n. If it converges the limit will be positive and finite for either scenario regardless of order (the limits will be different though). If it diverges then the limit will be 0 or infinity. 0 is not considered finite.

[Mathispoweru]

It converges due to the geometric series test |r| = 1/7 < 1.

[sakuranbo]

so what happens if the limit is 0 or infinity? my textbook say nothing about it but some websites say if the limit is 0, both converges, and both diverges at infinity

[liangpingshen]

@1:05, is it we can use the limit comparison test only if the direct comparison test doesn't work or we could also use it even if we know that the direct comparison test would work?