Calculus II - 9.2.2 The Geometric Series

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  • Опубліковано 17 гру 2024

КОМЕНТАРІ • 7

  • @夜听-g9w
    @夜听-g9w Рік тому

    this is so far the clearest explanation I have seen! Thank you so much, this saves my! 😊

  • @JoseAlvarado-ur5st
    @JoseAlvarado-ur5st 2 роки тому +8

    at 10:00, if it's (-1/3)^(n-1), then wouldn't you split that into (-1/3)^n * (-1/3)^-1, which in turn would make the second value -3?

  • @violintegral
    @violintegral 3 роки тому +6

    10:00 the series should be from 0 to inf of 3(-1/3)^(n+1), not ^(n-1). But the parts after that were still correct. Great video nonetheless. Keep it up!

    • @caseymiltner4991
      @caseymiltner4991 2 роки тому

      why?

    • @pianodan1608
      @pianodan1608 2 роки тому +2

      @@caseymiltner4991 because, if you plug in 0 for the power as n - 1, you get the first term to have a power of 0-1=-1 and not 1, which was what the original series gave you.

  • @onlynoone6512
    @onlynoone6512 3 роки тому +1

    5:50 but it isn`t geometric series? or it is?

    • @onlynoone6512
      @onlynoone6512 3 роки тому +1

      because n goes 1 to infinite🙄