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Math Help - Subsequences help, please!

  1. #1
    LadyCheese
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    Subsequences help, please!

    There is a sequence, <Xn>, such that the subsequence of even subscripts and odd subscripts both converge to x.

    How would I show, using this, that the whole sequence <Xn> converges to x?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by LadyCheese View Post
    There is a sequence, <Xn>, such that the subsequence of even subscripts and odd subscripts both converge to x.

    How would I show, using this, that the whole sequence <Xn> converges to x?
    well, the sequence \{ x_n \} is consists of terms of even and odd subscripts only. so saying the even and odd subscripts converge, is saying that every term of the sequence converges. but, for the sake of being more formal, we could do this. i'll start you off

    Let \{ x_n \} be a sequence. we denote the terms with even subscripts as x_{2k} and those with odd subscripts as x_{2k + 1}. if we let m = 2k, then the even terms are x_m and the odd terms are x_{m + 1}

    Now, since x_m \to x we have that for all \epsilon > 0 there exists an N_1 \in \mathbb{N} such that m \ge N_1 implies |x_m - x|< \frac {\epsilon}2. Similarly, for some N_2 \in \mathbb{N}, we have m + 1 \ge N_2 implies |x_{m + 1} - x|< \frac {\epsilon}2

    Now take such an \epsilon and choose N = \mbox{max} \{ N_1, N_2 \}. Then m, m + 1 \ge N implies that:

    |x_m - x_{m + 1}| = |x_m - x + x - x_{m + 1}| \le |x_m - x| + |x_{m + 1} - x| < \frac {\epsilon}2 + \frac {\epsilon}2 = \epsilon

    That is, we have |x_m - x_{m + 1}| < \epsilon



    Now, how would you continue. (we want to show that the sequence x_n converges. if that happens, then all its subsequences converge to the same limit as the sequence, which in this case is x. Hint: try to show x_n is a Cauchy sequence)
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