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Thread: 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.

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

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

    As $\displaystyle b_n $ converges to a in both cases, then $\displaystyle 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 $\displaystyle b_n $ converges to a in both cases, then $\displaystyle 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 $\displaystyle \epsilon >0, \exists N$ such that $\displaystyle \forall n\geq N$, you have$\displaystyle \lvert b_n-a\rvert < \epsilon$. You can find $\displaystyle N$ from the $\displaystyle N$ you would need for $\displaystyle a_n$. Thus $\displaystyle \lim b_n=a$.
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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 $\displaystyle \lvert a_n-a\rvert < \epsilon $ where $\displaystyle 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 $\displaystyle \lvert a_n-a\rvert < \epsilon $ where $\displaystyle n > \epsilon $.
    Since $\displaystyle a_n \to a$, for any $\displaystyle \epsilon >0$ there exists $\displaystyle n_0 \in \mathbb{N}$ such that $\displaystyle n>n_0 \Rightarrow|a_n-a|<\epsilon$. You need to show that the same $\displaystyle n_0$ works for the sequence $\displaystyle b_n$.
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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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