Proof: Sequence 1/sqrt(n) Converges to 0 | Real Analysis

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  • čas přidán 26. 08. 2024
  • We will prove the sequence 1/sqrt(n) converges to 0. In other words, we're proving that the limit of 1/sqrt(n) as n approaches infinity is 0. We use the epsilon definition of a convergent sequence and the proof is straightforward, following the typical form of a convergent sequence proof.
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Komentáře • 21

  • @oorischubert4085
    @oorischubert4085 Před 6 měsíci +1

    thanks great explanation

  • @SumanthGolagani
    @SumanthGolagani Před 11 dny

    Bro is there any way to use sandwich theorem for this?

  • @baffourawuah5676
    @baffourawuah5676 Před 22 dny

    easy to understand

    • @WrathofMath
      @WrathofMath  Před 21 dnem

      Glad to hear it, thanks for watching!

  • @Koj4
    @Koj4 Před 3 lety +1

    Great video. What are you plans for when you finish your Real Analysis playlist?

    • @WrathofMath
      @WrathofMath  Před 3 lety +4

      Thank you! Maybe abstract algebra, or I might really get going on a standard Calculus I playlist.

    • @Koj4
      @Koj4 Před 3 lety +1

      @@WrathofMath Oh, Abstract Algebra is nice. Great job man, your channel is going places.

    • @WrathofMath
      @WrathofMath  Před 3 lety +3

      It sure is! Thanks for your support! Down the line, I hope to make more traditional lectures on these subjects, so rather than a bunch of standalone videos on the topic arranged in a playlist, there would be a smaller collection of longer lectures - all cohesive and tied together as you would get from attending a course. But that's very demanding, so I'll have to grow the channel more before I can afford the time!

  • @benniefaithochimetiyi5134

    Thank you for the this video

  • @mahmoudalbahar1641
    @mahmoudalbahar1641 Před 3 lety +3

    Thank you for this good video...keep going with good work.

  • @tommasoc.2207
    @tommasoc.2207 Před měsícem

    in another video you made you set N = instead of N >, is there any difference? for example, if here we set N = 1/epsilon^2 instead of N > 1/epsilon^2, would things change?

    • @WrathofMath
      @WrathofMath  Před měsícem

      Generally speaking, no. There just needs to be some N that works. So whether we say "that N exists past this number" or "that N is this number", the result is demonstrated. Sometimes it is more convenient or clear to specify how big N needs to be (with inequality) than to give a precise expression that works with equality.

    • @tommasoc.2207
      @tommasoc.2207 Před měsícem

      @@WrathofMath thank you very much!

  • @alexiadrey4493
    @alexiadrey4493 Před 2 lety +1

    Amazing video once again! May I ask what the name of the song is at the end?

    • @WrathofMath
      @WrathofMath  Před 2 lety +1

      Thank you Alexia and I am glad you asked! At the end of my lessons, I often include music from my favorite songwriter who I wish more people knew about! That song is A Mask's Drawn smile, here is a link: crayonangel.bandcamp.com/track/a-masks-drawn-smile

    • @alexiadrey4493
      @alexiadrey4493 Před 2 lety

      @@WrathofMath Thank you for the info!!

  • @EhsanAmini
    @EhsanAmini Před 10 měsíci

    Thanks for the great videos. In the proof, is it ok to start with the inequality (n > an expression involving ε which we found in our scratch work) and work towards building |{aₙ} - L| < ε by manipulating both sides? E.g. in this case could we start by assuming n > 1/ε² ⇒ 1/n < ε² ⇒ 1/√n̅ < ε ⇒ | 1/√n̅ - 0 | < ε. Would it always work?

    • @harrietgomez8507
      @harrietgomez8507 Před 4 měsíci

      😊😊😊😊😊😊😊😊😊😊😊😊😊😊

  • @wiathamad3571
    @wiathamad3571 Před rokem

    Which app u were used please?