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Math Help - Sequences

  1. #1
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    Sequences

    Hi I am having a bit of trouble expressing the answer to a problem. If it is any constellation this problem goes alongside convergence proofs.

     Let \{a_n \}^\infty_1 , \{b_n\}^\infty_1 be sequences such that  lima_n = a and  b_n = a_n if n is even
    or  b_n = a if n is odd. Show that  limb_n = a

    That being said if n is even  b_{2n} = a_2 + a_4 + ... + a_{2n} . As  a_{2n} is a subsequense of  a_n one can conclude that  a_{2n} converges to  a . Hence  b_n will converge to a if n is even.
    As  b_{2n+1} = a is a is a constant.  b_{2n+1} will also converge to  a .

    As  b_n converges to a in both cases, then  limb_n = a as stated.

    I think this is the idea behind the question, if I overlooked anything please tell me. Thank you.
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  2. #2
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    Quote Originally Posted by Webby View Post
    Hi I am having a bit of trouble expressing the answer to a problem. If it is any constellation this problem goes alongside convergence proofs.
    You mean consolation.
    Quote Originally Posted by Webby View Post
    As  b_n converges to a in both cases, then  limb_n = a as stated.

    I think this is the idea behind the question, if I overlooked anything please tell me. Thank you.
    If you want to prove convergence of a sequence by subsequences, you have to show that every subsequence (not just the two you have selected) converges to the same number. An alternate route is to show that given \epsilon >0, \exists N such that \forall n\geq N, you have  \lvert b_n-a\rvert < \epsilon. You can find N from the N you would need for a_n. Thus \lim b_n=a.
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  3. #3
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    This may sound like a silly question but how would you find n>N in this case  \lvert a_n-a\rvert < \epsilon where  n > \epsilon .
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  4. #4
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    Quote Originally Posted by Webby View Post
    This may sound like a silly question but how would you find n>N in this case  \lvert a_n-a\rvert < \epsilon where  n > \epsilon .
    Since a_n \to a, for any \epsilon >0 there exists n_0 \in \mathbb{N} such that n>n_0 \Rightarrow|a_n-a|<\epsilon. You need to show that the same n_0 works for the sequence b_n.
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  5. #5
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    can i say that let N = max{ 2no, 2no +1 } then for all n> N, bn converges to a?
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