Results 1 to 2 of 2

Math Help - Sequence Question (not too hard)

  1. #1
    Senior Member slevvio's Avatar
    Joined
    Oct 2007
    Posts
    347

    Sequence Question (not too hard)

    Hello fellow mathematicians I hope that the summer hasn't rusted your brains here is a question:


    Let  \{ x_n \} be a sequence and  L \in \mathbb{R} .
    Suppose that  \{ x_{2n} \} and  \{ x_{2n-1} \} both converge to  L .
    Prove that  \{ x_n \} also converges to L.

    Is this the correct way of doing this? I'm quite rusty at the moment and any help would be appreciated.

    We know the following to be true:

    •  \forall \epsilon > 0 \exists N_1 > 0 such that  n > N_1 \implies |x_{2n} - L | < \epsilon
    •  \forall \epsilon > 0 \exists N_2 > 0 such that  n > N_2 \implies |x_{2n-1} - L | < \epsilon
    Let  u = 2n and  v = 2n - 1 .  u,v \in \mathbb{N} since  2n > 0, 2n - 1 > 0 \forall n \in \mathbb{N}
    So we get:

    •  \forall \epsilon > 0 \exists N_1 > 0 such that  \frac{u}{2} > N_1 \implies |x_{u} - L | < \epsilon
    •  \forall \epsilon > 0 \exists N_2 > 0 such that  \frac{v+1}{2} > N_2 \implies |x_{v} - L | < \epsilon
    which after replacing variables and moving the inequality, is equivalent to :


    •  \forall \epsilon > 0 \exists N_1 > 0 such that  n > 2N_1 \implies |x_{n} - L | < \epsilon
    •  \forall \epsilon > 0 \exists N_2 > 0 such that  n > 2N_2 - 1 \implies |x_{n} - L | < \epsilon
    Let  \epsilon > 0 . Choose  N = \max \{2N_1,2N_2 - 1\} . Let  n \in \mathbb{N} .  n > N \implies |x_n - L| < \epsilon .
    Is this proof any good? I don't really know I kind of just scribbled it down but if there is anythign wrong I would like to know thanks for the help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Yes, that looks completely correct.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sequence Proof (Hard)
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: October 7th 2009, 12:20 AM
  2. This question is so hard!! any help??
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 4th 2008, 11:59 AM
  3. sequence, not easy but not hard
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: December 9th 2007, 12:08 PM
  4. Induction, sequence, not hard, please help me
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 19th 2007, 11:57 AM
  5. Need some help with a hard question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 23rd 2007, 05:05 AM

Search Tags


/mathhelpforum @mathhelpforum