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Thread: Convergence of sequence

  1. #1
    Nov 2008

    Convergence of sequence

    I'm grateful to any help.
    I'm supposed to show that $\displaystyle x_{n}-x_{n-1} \rightarrow 0$ as $\displaystyle n \rightarrow \infty$ if $\displaystyle \{x_n\}$ converges.
    Suppose the sequence converges to x. Since $\displaystyle \{x_n\}$ converges, $\displaystyle \{x_{n-1}\}$ also converges.
    This implies that for every $\displaystyle \epsilon \textgreater 0$, there exist integers $\displaystyle N_1, N_2$ s.t $\displaystyle N_1 \textgreater n$ and $\displaystyle N_2 \textgreater n-1$ implying absolute value of $\displaystyle (x_n-x) \textless \epsilon$ and absolute value of $\displaystyle x_{n-1}-x \textless \epsilon$.
    Am I on the right track to solve this problem? Here, do I need to use the definition of limit to show that $\displaystyle lim (b_n -b_{n-1})=0$ as $\displaystyle n \rightarrow \infty$
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  2. #2
    Super Member girdav's Avatar
    Jul 2009
    Rouen, France
    We have $\displaystyle \left|x_n-x_{n-1}\right| = \left|x_n-l+l-x_{n-1}\right| \leq \left|x_n-l\right|+\left|x_{n-1}-l\right| \leq \frac{\epsilon}2+\frac{\epsilon}2$ for $\displaystyle n \geq \max\left(N_1,N_2\right)$.
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