Results 1 to 2 of 2

Thread: proof for calc 5

  1. #1
    Junior Member
    Dec 2006

    proof for calc 5

    Suppose that {a_n} is a divergent sequence. Show that it has a subsequence {a_{n_k}} with 1+ n_k < n_{k+1} which also diverges

    can anyone give me a concise proof for this
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    You could try an argument by contradiction. Suppose the result is false, so that every such subsequence converges. In particular, the subsequences a_{2n} and a_{2n+1} (of even-numbered and odd-numbered terms terms in the sequence) both converge, say a_{2n}\to l_1 and a_{2n+1}\to l_2. The subsequence a_{3n} also converges, and it's easy to see that its limit has to be both l_1 and l_2. So that means that l_1=l_2, and the whole sequence has to converge to that limit.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A few Pre-Calc problems in my AP Calc summer packet
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Aug 30th 2010, 04:40 PM
  2. Replies: 1
    Last Post: Jan 13th 2010, 12:57 PM
  3. Multivariable Calc Chain Rule proof problem
    Posted in the Calculus Forum
    Replies: 7
    Last Post: May 27th 2009, 09:13 PM
  4. calc 3 proof
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jun 3rd 2008, 01:32 PM
  5. Replies: 2
    Last Post: May 14th 2008, 08:56 AM

Search Tags

/mathhelpforum @mathhelpforum