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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 x_{n}-x_{n-1} \rightarrow 0 as n \rightarrow \infty if \{x_n\} converges.
    Suppose the sequence converges to x. Since \{x_n\} converges, \{x_{n-1}\} also converges.
    This implies that for every \epsilon \textgreater 0, there exist integers N_1, N_2 s.t N_1 \textgreater n and N_2 \textgreater n-1 implying absolute value of (x_n-x) \textless \epsilon and absolute value of 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 lim (b_n -b_{n-1})=0 as n \rightarrow \infty
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  2. #2
    Super Member girdav's Avatar
    Jul 2009
    Rouen, France
    We have \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 n \geq \max\left(N_1,N_2\right).
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