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Homework Help: Limit of a sequence

  1. Mar 7, 2005 #1
    i had this problem in my book that i wasn't able to do. I kinda had the idea of what it involved but just wanted to clear it up with you guys.
    So the problem is:
    Suppose that an ->L and bn ->L. Show that a1,b1,a2,b2,... converges to L.

    So here it seems to me like i can obviously define a new sequence cn, like that in the pinching theorem. But using other definitions(epsilon, etc.) of sequences, how do i come up with a proof of this.
  2. jcsd
  3. Mar 8, 2005 #2
    Fix an epsilon. That a_n and b_n converge to L implies there is some N (take each sequence individually, choose an M and M' for each, take the bigger of the two and double that) such that whenever n>N both a_(n/2) and b_(n/2) are within epsilon of L (and therefore c_n is within epsilon of L).
    Last edited: Mar 8, 2005
  4. Mar 8, 2005 #3
    could someone check this proof for me and tell me what is missing as i m not sure if i know anymore:(problem stated above)
    If a_n ->L and b_n ->L for some L, then for any eps>0 there is a K such that for all n, if n>K, then |a_n - L|<eps/2 and |L - b_n|<eps/2.
    Since n>K, |a_n - b_n|</= |a_n - L| + |L - b_n|<eps/2 + eps/2 =eps.
    (now to show that it is bounded)
    If eps=1, then |a_n - b_n|<1 for some n>K.
    This means |a_n - b_K+1|<1 for all n>K.
    Thus {a_n:n>K} and {b_n:n>K} are bounded.
    So a_n ->L and b_n->L.
  5. Mar 8, 2005 #4
    I don't think you have to show anything is bounded. That will follow automatically once you prove the limit is L. What you want to do is define the sequence by [tex]\{c_n\}_{n\in \mathbb{N}}[/tex] by [tex]c_n=a_{[(n+1)/2]}[/tex] if n is odd and [tex]c_n=b_{(n/2)}[/tex] if n is even. Now fix an epsilon > 0. There are N and M such that whenever n>N, a_(n+1)/2 is within epsilon of L and whenever n>M b_(n/2) is within epsilon of L. Take the bigger of the two, let's say it is P. Then whenever n>P, both a_(n+1)/2 and b_(n/2) are within epsilon of L. That is, whenever n>P, c_n is within epsilon of L.
    (sorry for the sloppier version of this above)
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