Limit comparison test | Series | AP Calculus BC | Khan Academy

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  • Опубліковано 19 жов 2024

КОМЕНТАРІ • 46

  • @MauricioMartinez0707
    @MauricioMartinez0707 7 років тому +45

    Jesus man you're fucking awesome. Theres a lot of other good teachers on youtube, like professor leonard, and partcikjmt and such, but the one thing about your teaching is that you get the point across quickly. I dont have to spend 2.5 hours watching a whole lecture, or 6-7 different videos. You give us the theorems, maybe an example here and there, in like a 6 minute video. Thanks khan academy. Ya'll know whats up

  • @yewut
    @yewut 8 років тому +62

    take my tuition

  • @vee361-e1c
    @vee361-e1c 8 років тому +6

    Thanks a bunch!! Really well done, i understood direct comparison test but LCT flew over my head until now.

  • @Anastasia231
    @Anastasia231 8 років тому +4

    Thank you sir! this helped lots!

  • @tejasgajra2731
    @tejasgajra2731 7 років тому +1

    If lim (an/bn) = t (t>1)
    Then an > bn ... If bn converges then it is not necessary for "an" to. Converge...
    Pls can you prove the theorem?

  • @chaloondolaro6732
    @chaloondolaro6732 3 роки тому

    Your voice is so calming 🤍

  • @walkieer
    @walkieer 4 роки тому

    For the second example can you use the traditional comparison test and say 1/1^n >= 1/(2^n - 1). Or is there something wrong there?

  • @HighStar9821
    @HighStar9821 7 років тому +2

    are there more specific requirements for bn?
    like if an is 1/2^n can i just set bn at 100^n?

  • @rashmikababajee3328
    @rashmikababajee3328 6 років тому +1

    helloo,
    your tutorials are amaazzinnggg :D
    the only thing that I fail to understand is when do we know when to put n=1 or n=0 below the sigma notation...
    Can somebody please explain it to me :)
    Thanks in advance:)

    • @haidarshehade241
      @haidarshehade241 3 роки тому

      Sorry for being late but it's a given. You don't really put it. But n determines the shape of the series to determine if it's geometric or to define a mathematical function or number like eˣ or sin(x) or ln2 or whatever to be determined, which all depends on a specific n to start with.

  • @angelakim4085
    @angelakim4085 8 років тому +1

    Thank you so much !

  • @dikshasabharwal3754
    @dikshasabharwal3754 6 років тому +1

    Thanks.😄

  • @carlmoller807
    @carlmoller807 9 років тому +1

    You're awesome dude!

  • @miraix6493
    @miraix6493 8 років тому +1

    thank you

  • @nataliecross9403
    @nataliecross9403 7 років тому +5

    How do we know 1/2^n converges?

    • @theerancheliyan6254
      @theerancheliyan6254 7 років тому +16

      1/2^n = 1^n/2^n = (1/2)^n and since 1/2 is greater then -1 and less then 1 it converges via geometric series. Hope this helps.

    • @itecnus3490
      @itecnus3490 7 років тому +3

      I like how you've got 86 subs for having a cute pic.

    • @sssingh6848
      @sssingh6848 6 років тому

      Natalie Cross
      Its because when u take limit as n tends to infinity then it will give a unique finite value

  • @VickiBee
    @VickiBee 10 років тому

    What is this called? Calculus?
    The doctor treating me now says I have a tumor on the right side of my brain but that makes no sense because I've never felt pain on the right side, only on the left. He's got me all confused. I've always been left-handed and used my right brain side more than the left and now he's telling me they found a hemangioma on the right side.
    When you do math you use the left side of your brain, which always hurts when I use it too much. I've never felt a headache on the right side of my head, only the left. That's what makes me think is the reason I'm so lousy at math.

  • @yashraj4272
    @yashraj4272 6 років тому

    Plz give homework questions at the end of your videos plzzz

  • @mus1cal4ddict76
    @mus1cal4ddict76 4 роки тому

    So one use the limit comparison test when the direct comparison test gives you and inconclusive answers????

  • @90kyro
    @90kyro 7 років тому

    But (1/2^n) doesn't Converge, it Diverges and is a Harmonic Series right?

    • @alaguthiagarajan4139
      @alaguthiagarajan4139 7 років тому

      it converges. as n grows larger the denominater grows exponentally from 1/2 to 1/4 to 1/8 and so on. The numbers get smaller and at infinity should approach 0. It doesn't diverge for to diverge it must be unbound and it clearly isn't for it can only grow smaller towards 0.

    • @enriquerosas4510
      @enriquerosas4510 7 років тому

      Harmonic series = 1/n, 1/2^n is geometric series

    • @tejasgajra2731
      @tejasgajra2731 7 років тому

      Nah.. It converges, actually its a geometric series with common ratio=1/2 and G.S converges if |r|

  • @moose7145
    @moose7145 5 років тому

    But what if the limit is zero when you do lim comparison. It's not positive??

    • @robertohernandez1642
      @robertohernandez1642 5 років тому

      Theorem definition says it HAS to be greater than zero, so if its zero then they dont act the same, but different.

  • @shayorshayorshayor
    @shayorshayorshayor 4 роки тому

    What if the limit comparison test equals 0?

  • @MrBoriskaful
    @MrBoriskaful 7 років тому +1

    I dont know when to use which method AHHH my exam is on saturday

  • @uduak-abasiamana9914
    @uduak-abasiamana9914 8 років тому

    what does he mean by positive and finite?

    • @MrIrvydas
      @MrIrvydas 8 років тому +3

      +AfricannBaabeeess any number that doesn't have - in front of it, but not infinity.

    • @iimousa97
      @iimousa97 7 років тому

      how do we choose an and bn?

    • @haidarshehade241
      @haidarshehade241 3 роки тому

      A positive non zero constant

  • @joshuachamblee8499
    @joshuachamblee8499 9 років тому

    Why did he divide the numerator and the denominator by 2^n at 5:45

    • @kuaana94
      @kuaana94 9 років тому +1

      Joshua Chamblee Dividing by a fraction is the same as multiplying by its reciprocal. He had 1/2^n in the denominator so he just rearranged it to be 2^n in the numerator. This then becomes 2^n * 1/(2^n-1) so 2^n-1 ends up in the denominator. Hope that helps!

    • @vee361-e1c
      @vee361-e1c 8 років тому

      I know this is old, but Joshua was asking why he divided the whole 2n/(2n-1) by 2n. It's to simplify it so it's not an indeterminate form and a finite limit can be found.

  • @sangamesh2727
    @sangamesh2727 2 роки тому

  • @bageoop
    @bageoop 10 років тому +4

    Khan Academy has 1 million subs and this video has 0 comments? huh

  • @dereksavage8728
    @dereksavage8728 2 роки тому

    Thank you :_)

  • @HL-iw1du
    @HL-iw1du 6 років тому

    d

  • @prabodhbharose4209
    @prabodhbharose4209 5 років тому

    1.25 speed