Absolute Convergence, Conditional Convergence, and Divergence

[@TheOrganicChemistryTutor]

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Professor Organic Chemistry Tutor, thank you for a fantastic explanation of Absolute Convergence, Conditional Convergence and Divergence in Calculus Two. The examples in this video, increases my full understanding of this topic in Calculus. This is an error free video/lecture on KZbin TV with the Organic Chemistry Tutor.

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[@LuminousOriens]

For the last problem, when checking 'a sub n' after getting that '|a sub n|' was divergent, why would you include the (-1)^n when finding the limit for the alternating series? shouldn't we just try to find whether 'a sub n+1' is smaller than 'a sub n' and exclude the part that's changing the signs?

[@fryPS]

I’m so confused on that part too. And for the absolute value expression, why did he apply the alternating series test instead of the comparison test since the sign is gone?

[@srmback]

I think tutor was confused but anyway, in last task you get 1 from limit in both cases that means it is automatically divergent by checking Alternating Series Test (does not passes 1st test).

[@LuisBiland]

I think he uses the alternating series test because it would be more efficient to test whether the series converges or diverges rather than taking a direct comparison, since the problem is already set up nicely for taking the alternating series test

[@abhinavgulati6169]

Yeah because the answer of the last part would be conditionally convergent not divergent.

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[@corinnehentschel6103]

Great video. Needed a bit of extra help. Thank you!

[@ardakral5973]

11:41 don't we have to take the limit of a(n) only without (-1)^n included ?

[@camalmuradov274]

yes, I also think so

[@ilievskiviktor]

Yes, but here the limit is still not zero. However keep in mind to get rid of (-1)^n or (-1)^n+1 because it does not belong in An.

[@Brantendo64]

I'm confused with this as well. Of course anything with (-1)^n multiplied it isn't going to converge... That's why we have the alternating series test. Why is it that in this particular case we can just skip the test altogether? This is by far the worst part of calculus. Most of it isn't consistent. You have to rely on memory rather than math skill.

[@sagniksaha7359]

Hey you would do that (and you would be absolutely correct) if you were doing the ALTERNATING SERIES TEST. Here he used the DIVERGENCE TEST. In this case, either of those tests will work. Hope that clears it up.

[@rayzhangelite]

I think it's because the initial limit test for An did not come up to be 0 and therefore it failed the alternate series test and he had to use divergence test.

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[@bman9469]

11:30 Why is the (-1)^n not removed from the a_n for the alternating series test like the previous example?

[@rockieRAGE117]

because he was doing the divergence test instead. The alternating series test is failed.

[@arisoda]

@@rockieRAGE117 But the Divergence Test is part of the Alternating Series Test... How does that work?

[@shayorshayorshayor]

arisoda - so try doing the alternating test anyways (on n^3 / n^3 +5 ) it’s going to fail. So he skipped the alternating test anyways and just went straight on doing the divergent test. He shouldn’t have done it like that because it confused me as well. The previous example would’ve have failed if he skipped the alternating test. So I get your confusion.

[@arisoda]

Faisal Kabir But how can one SKIP the alternating series test if the divergence test is part of the alternating series test and then go ahead do the divergence test. Isn't he still doing the alternating series test then anyway?

[@assassinmaster8044]

11:27 you wrongly took (-1)^n in the lim an. Although it does not affect the final answer as lim would be 1 which is not equal to 0.

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[@eulzzzz7682]

Why are you leaving the abs for others except for(-1)^n???

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[@jackien404]

Why isn't it necessary to prove convergence for the first question using limits like you do in the second one if they are both alternating series?

[@prettylittlelife3735]

i think it í because we checked the series with absolute value mark and saw it is convergence => the original is convergent too.

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How would you solve a problem that has nsquareroot(n) instead of it being being 3squareroot(n)?

[@immadshaikh9640]

why during the p-series test, the 1/[(3)^1/3] diverges but during Lim->∞ 1/[(3)^1/3]=0 and converges?? Please explain, I am confused!!

[@jamesxia9188]

I have the same question, did you ever find an answer?

[@LongThanh-iv6ih]

​@@jamesxia9188 because /a_n/= 1/n^1/3 is the p-series so when the p is less than 1, it div. But in the other hand, the series a_n is an alternating series so he must use 2 condition to test whether this seris conv or not. The 1st conditon is the /a_n+1/ is less than or equals /a_n/ ( /a_n/ i mean the absolute value of a_n). The 2nd one is the lim of a_n (or /a_n/) when n comes to infi, equals to 0. Hope my explanation can help you ^^

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[@alejandroarreguin3628]

For the last question why didn't you use AST for a_n to determine if it converged or diverged?

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[@MahadSalman]

Brother at 6:21 you said p test n is less than one but how it diverge it converges as we put n from 1 to infinity?

[@anniecheng1600]

I could be wrong but I think the p test showed that the absolute value of a_n diverges. But as he stated at the beginning of the video we still need to test the original series to see if it's conditionally convergent or divergent.

[@MahadSalman]

That makes sense 👍

[@alvinbensongocheco612]

Why isn't it An - 1 7:45? cause its the term before the last term?

[@monkeyking10192003]

How come some times you include the (-1)^n in the a_n = statement, but other times you leave it out?

[@DaHoKilla]

yeah I dont get that either, i thought you always ignore it for AST and ABS V test

[@mtinfinity7845]

Thank you so much you're great but I'm still gonna fail my sequences and series exam because all this info is way too much for me to memorize...

[@redmer_de_boer]

Im confused about when to take a_n and when to take the entire series with (-1)^n included. In your video you take a_n to be everything except (-1)^n, but if thats the case with an alternating series, then there's never any difference between the normal series and the absolute value of the series since you take away the thing that causes it to be alternating. How does this work?

[@eatingourassesoff]

The Alternating series test checks for three things: 1. That the series is alternating (-1)^n for example, 2. that the portion multiplying the -1^n is "bounded" that is that it is going to eventually stop growing (that's the limit=0), and 3. That the rate of change (or each subsequent term) is smaller than the previous (monotonically decreasing). If something is bounded and decreasing then it doesn't matter if the (-1)^n is there since it is adding/subtracting less and less each time towards an end and thus we can ignore the (-1)^n portion.

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Thank you!

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[@AyomideAdebowale-rd6gd]

please can you explain the convergence and divergence of improper integral

[@nj-j7034]

This video is pretty clear. However, my university textbook is like a completely different one.

[@凛-r3h]

7:30 Here you said "it(a_n) passes the divergence test", this confused me. Then after you found a_(n+1)

[@josephgeorge1785]

Bro ig divergence test is another name for Leibnitz test. So what you do in Leibnitz test is that you check if the series is monotonically decreasing and if the limit of the series excluding the alternating part ,i.e. (-1)^n is 0. If both of these conditions satisfy then we conclude that the series or sequence is convergent

[@bonelesspizza6311]

What about the series (-1)^n ? Its absolute value converges but the series diverges.

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do we solve (cos n) the same way as (cos n.pi)?

[@jeyhuntalibov9911]

Thank you so much sir. One thing i haven't get it is how can i know when limit doesn't exist? is there any rule and what is that rule? Thanks beforehand

[@tafadzwamachuwe4118]

How come in the last example we didn’t remove (-1)^n when using divergent test as we did with the second example.

[@blackpinktanked6809]

You only remove (-1)^n when using an alternating series test which is only applicable for summations, and not limits

[@prevailtm]

Not sure if anybody else was confused about this like I was, but if you have 1/cube root n, can't you rewrite it as 1/n^(1/3)? He wrote that earlier and explained how it was divergent with p-series, and then rewrote it and all of a sudden it's equal to zero?

[@jacobq49]

exactly what im confused on

[@adeyemoayomikunadewole4984]

You're too good

[@raghavendaranagavarma3003]

thank u so much bro and i have one doubt if mod an is converent and an is divergent ,what is the answer

[@GenoRL_]

how come at 7:30 you didn't just use a p-series again? also at 11:33 I thought the (-1)^n disappeared when doing the alternating series test like it did in the example at 7:16. There is some tomfoolery going on here and I don't like it.

[@blackpinktanked6809]

7:30 You only use the p-series when finding the convergence of a summation of 1/n^p and not the limit. 11:33 Again, you're finding a limit. There is no summation involved so you don't use the alternating series test (which use if there is a summation sign)

[@Princessa98]

hi by means of having already found the absolute value of the series to be divergent, wouldn't that automatically mean the series itself is convergent making it conditionally divergent?

[@3k04t9]

To anyone seeing this 4 years later- think about it like squares and rectangles. Every square is a rectangle, but not all rectangles are squares. If a series is ABSOLUTELY convergent, then it is like a square, in the sense that is both a rectangle and a square. If we know a series is conditionally convergent, then it is like a rectangle. It is possible for it to be a square as well, or maybe it's not. That's what checking the absolute value of the series determines - whether it is a square or just a rectangle. Finally, if a series isn't conditionally convergent (keeping up with the example, imagine a different shape entirely, like a circle) then we know, by default, it is also not absolutely convergent (and therefore also not a square). And if a series isn't convergent, it must be divergent.

[@benslavik582]

What if they tell you to find absolute convergence and there is an X or (-X)^n in the series

[@vaniabessos6381]

very helpful, thank you

[@arthurstephenson869]

im just neurodivergent

[@nicoleyravioli6721]

Wait why are we comparing to the original and not an? We did that for the alternating series. Why did we change it up?

[@Viper1924]

The original is the an. Alternating series just deals with the an, but to find if something is conditionally or absolutely convergent, we must find if it converges or diverges for the absolute value of |an| and then find if the original (an) converges or diverges using alternating series.

[@miguelcalles4795]

Nice video! Thanks!

[@prettylittlelife3735]

in the last task, why lim |an| = 1 then it diverges? i think it is convergent to 1

[@af2954]

diverges through divergence test. but yeah it's supposed to be conv I think, he used AST for the abs cone test when that would be used to test conditionally cone

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Thanks to you man

[@user-nl6vu8zo1d]

Thank you

[@Dina-he1uc]

I think in example #3 you cannot use the alternating series test because it is an increasing function (n^3/(n^3+5))

[@christianwang]

its decreasing because the denominator is larger than the numerator

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That's not how you determine that smh

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We're almost done with Calculus 2! Lets goooooo

[@wamo693]

I have a question. If we find that the absolute series is convergent, then we don't have to find the convergence for the given series? Would be really helpful if anyone could answer for me.

[@Michael-hs3if]

you still have to, "absolute convergence" means that is is convergent on both the given series and absolute series

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Thank you for your response!

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You don't have to as he said in the video ,hope this reply ain't too late

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Knowing all this still gets me a bad grade because our Calc professor's homework and quizes never match the actual test material. Its always trick problems that I have never seen.

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[@breannag3055]

don’t know if anyone will answer but in a problem: { (-1)^n • 1/n (pretend { is the summation symbol from 1 to infinity) i would before use the alternating series test to find it to be convergent, but now i’m learning what was just shown in this video so with that same problem it would be conditionally convergent sooo is the alternating series test just a test to use to help prove it is conditionally convergent? this might be a dumb question but should the method in this video always be used first before jumping into the alternating series test?

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breanna g i think the alternating series would just be used

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Good night