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Math Help - Convergence of a sequence

  1. #1
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    Acolman, Mexico
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    Convergence of a sequence

    Hello,

    I am stuck with this question.

    Suppose x_n \leq 0 for all n \in \mathbb{N} and that \lim ((-1)^n x_n) exists. Show that (x_n) converges.

    Ok, here's what I done so far...
    If \lim ((-1)^n x_n) exists, then ((-1)^n x_n)is a convergent. Thus, a Cauchy sequence.
    So there exists a M such that for all \epsilon > 0, if n,m>M then |(-1)^m x_m - (-1)^n x_n| < \epsilon
    So, if we assume that n > M, then
    |(-1)^{n+1} x_{n+1} - (-1)^n x_n| < \epsilon
    \Rightarrow |(-1)^n ( -x_{n+1} -  x_n)| < \epsilon
    \Rightarrow |(-1)^n||( -x_{n+1} -  x_n)| < \epsilon
    \Rightarrow |-x_{n+1} -  x_n| < \epsilon
    \Rightarrow |x_{n+1} +  x_n| < \epsilon
    Since, x_n \leq 0 for all n \in \mathbb{N}, then does that mean that x_n = 0 ??
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  2. #2
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    Quote Originally Posted by akolman View Post
    Hello, I am stuck with this question.
    Suppose x_n \leq 0 for all n \in \mathbb{N} and that \lim ((-1)^n x_n) exists. Show that (x_n) converges.
    Come on. Think about it.
    We know that \left| {\left| {y_n } \right| - \left| L \right|} \right| \leqslant \left| {y_n  - L} \right|.
    So it follows that y_n  \to L\, \Rightarrow \,\left| {y_n } \right| \to \left| L \right|.

    Can you apply that to this problem?
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