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Math Help - x2n , x2n+1 converges to L then xn converges to L

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    x2n , x2n+1 converges to L then xn converges to L

    I am trying to prove that if x2n converges to L and if x2n+1 converges to L then xn converges to L, however I have been having problem combine the two information and show it for xn.

    So we know that if x2n converges to L, for every epsilin there exists N1 such that for every n>N1 |x2n-L|<epsilon
    and for every epsilon there exists N2 such that for every n>N2 |x2n+1-L|<epsilon

    So if we define N=max{N1,N2} we will get that for every n>N |x2n-L|<epsilon and |x2n+1-L|<epsilon
    And therefor for every n>N we can say |xn-L|<epsilon? because our n is either odd or even, in both cases we get |xn-L|<epsilon .

    Is my logic correct? And is it formal? How can I make it more clear ? I think it's correct however coming to the conclusion that |xn-L|<epsiulon was too fast. It seems like I am missing a step . Appreciate if someone can give a feedback. Thank you.
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    Re: x2n , x2n+1 converges to L then xn converges to L

    Quote Originally Posted by davidciprut View Post
    I am trying to prove that if x2n converges to L and if x2n+1 converges to L then xn converges to L, however I have been having problem combine the two information and show it for xn.

    So we know that if x2n converges to L, for every epsilin there exists N1 such that for every n>N1 |x2n-L|<epsilon
    and for every epsilon there exists N2 such that for every n>N2 |x2n+1-L|<epsilon

    So if we define N=max{N1,N2} we will get that for every n>N |x2n-L|<epsilon and |x2n+1-L|<epsilon
    And therefor for every n>N we can say |xn-L|<epsilon? because our n is either odd or even, in both cases we get |xn-L|<epsilon .

    Is my logic correct? And is it formal? How can I make it more clear ? I think it's correct however coming to the conclusion that |xn-L|<epsiulon was too fast.
    No, I think it is fine as is.
    Thanks from davidciprut
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