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

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

    Can you apply that to this problem?
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