Results 1 to 5 of 5

Math Help - nth term test question.

  1. #1
    Senior Member
    Joined
    Jan 2007
    Posts
    477

    nth term test question.

    This problem diverges: lim n->infinity (k^2) / (k^2 -1)

    this diverges by the nth term test. =/ 0 the limit equals 1.

    however, my question is, since this equals 1. Why does this not converge to 1?

    is this because it doesn't converge to 1 quick enough?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by rcmango View Post
    This problem diverges: lim n->infinity (k^2) / (k^2 -1)

    this diverges by the nth term test. =/ 0 the limit equals 1.

    however, my question is, since this equals 1. Why does this not converge to 1?

    is this because it doesn't converge to 1 quick enough?
    It's simple to see that the SEQUENCE converges to 1. eg. \frac{k^2}{k^2 - 1} = 1 + \frac{1}{k^2 - 1}.

    The nth term test is used to test whether a SERIES, not a sequence, diverges!! And clearly the series \sum_{n = 2}^{\infty} \left (1 + \frac{1}{k^2 - 1} \right) diverges (the sum of 1's --> oo). Unsurprising since the nth term does not approach zero.

    Or are you asking for a proof of the nth term (limit) test? If so, consult any standard reference on the topic .....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1
    Also we have  \sum\limits_{k = 2}^\infty  1  \le \sum\limits_{k = 2}^\infty  {\left[ {1 + \frac{1}{{k^2  - 1}}} \right]} .
    Adding infinitely many ones diverges.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jan 2007
    Posts
    477
    Okay, didn't know it was testing a series, i see now. thanks.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2008
    Posts
    6
    it doesn't converge to one because in a geometric series, if a sub n doesn't equal 0, then it doesn't converge at all. Therefore, the sum might be 1, but that doesn't mean it converges to 1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: February 19th 2011, 12:21 PM
  2. Replies: 0
    Last Post: December 30th 2010, 10:36 AM
  3. is it divergent via nth term test?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 18th 2010, 09:47 PM
  4. Power series, term-by-term integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 8th 2010, 03:14 AM
  5. Calculus question regarding term
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 1st 2009, 10:42 AM

Search Tags


/mathhelpforum @mathhelpforum