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Posers and Puzzles

Posers and Puzzles

  1. 15 Oct '07 17:10
    Determine whether the following sequence converges or diverges. If it converges find the limit. an=n(sin(1/n))
  2. Standard member PBE6
    Bananarama
    15 Oct '07 17:29
    Originally posted by Arachnarchist
    Determine whether the following sequence converges or diverges. If it converges find the limit. an=n(sin(1/n))
    We can rearrange this sequence and make a simple substitution to make it more recognizable:

    a(n) = n*sin(1/n) = sin(1/n)/(1/n)

    Let x = 1/n. Then we have:

    a(x) = sin(x)/x

    As n->inf, x->0, so we must find lim(x->0) a(x). This is easily done using L'Hopital's rule:

    lim(x->0) sin(x)/x = lim(x->0) cos(x)/1 = 1

    So this sequence does converge to a value of 1.
  3. Standard member adam warlock
    Baby Gauss
    15 Oct '07 17:32 / 2 edits
    Originally posted by Arachnarchist
    Determine whether the following sequence converges or diverges. If it converges find the limit. an=n(sin(1/n))
    Well the limit of the general term when n tends to infinity is
    lim an = lim n*sin(1/n) = lim n/n = 1.
    Since the general term doesn't tend to zero we can say that the series diverges.

    Edit: Somehow I read your post as a question if the series of general term n*sin(1/n) converges. Sorry about that. Having read PB's answer I realised that I read it wrong.
  4. Standard member TheMaster37
    Kupikupopo!
    15 Oct '07 17:46
    Originally posted by Arachnarchist
    Determine whether the following sequence converges or diverges. If it converges find the limit. an=n(sin(1/n))
    As for n -> 0

    sin(1/n) is bounded, so n*sin(1/n) -> 0
  5. Standard member adam warlock
    Baby Gauss
    15 Oct '07 17:56
    Originally posted by TheMaster37
    As for n -> 0

    sin(1/n) is bounded, so n*sin(1/n) -> 0
    It's n*sin(1/n) not 1/n*sin(1/n).
    The first time I also read this wrong though not like you
  6. 15 Oct '07 18:29
    Originally posted by TheMaster37
    As for n -> 0

    sin(1/n) is bounded, so n*sin(1/n) -> 0
    I think he meant n -> infinity, not n -> 0. I'll go with what PBE6 said.
  7. 15 Oct '07 18:30
    Originally posted by adam warlock
    Edit: Somehow I read your post as a question if the series of general term n*sin(1/n) converges. Sorry about that. Having read PB's answer I realised that I read it wrong.
    Sequence v Series
  8. Standard member TheMaster37
    Kupikupopo!
    15 Oct '07 20:10
    Originally posted by adam warlock
    It's n*sin(1/n) not 1/n*sin(1/n).
    The first time I also read this wrong though not like you
    One thing you learn in time is to read carefully. It makes life so much easier at times

    Tohugh I occasionally muck up as well :p
  9. Standard member adam warlock
    Baby Gauss
    15 Oct '07 21:16
    Originally posted by TheMaster37
    One thing you learn in time is to read carefully. It makes life so much easier at times

    Tohugh I occasionally muck up as well :p
    LOL
    The thing is that I'm only used to calculating limits as n goes to infinity is this types of exercises. I think now that I didn't even read your post . Automod was on full power.
  10. Standard member genius
    Wayward Soul
    19 Oct '07 07:58
    Originally posted by Arachnarchist
    Determine whether the following sequence converges or diverges. If it converges find the limit. an=n(sin(1/n))
    why "calculus"?
  11. Standard member PBE6
    Bananarama
    19 Oct '07 14:11
    Originally posted by genius
    why "calculus"?
    Because you have to take the limit as "n" tends toward infinity to see if the sequence converges or not. That's some straight-up calculus, yo.
  12. Standard member genius
    Wayward Soul
    19 Oct '07 15:12
    Originally posted by PBE6
    Because you have to take the limit as "n" tends toward infinity to see if the sequence converges or not. That's some straight-up calculus, yo.
    *Bows his head in shame as he re-learns what he thought the Calculus was*
  13. Subscriber sonhouse
    Fast and Curious
    27 Oct '07 13:43
    Originally posted by PBE6
    Because you have to take the limit as "n" tends toward infinity to see if the sequence converges or not. That's some straight-up calculus, yo.
    What does PBE6 mean? You want 1,000,000 Pall Bearers?
  14. Subscriber AThousandYoung
    It's about respect
    27 Oct '07 18:04
    Originally posted by sonhouse
    What does PBE6 mean? You want 1,000,000 Pall Bearers?
    Pirate Banana Evildoer #6.

    http://www.hello-cthulhu.com/?date=2004-09-28