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Thread: Simple sequences problem

  1. #1
    Nov 2008

    Simple sequences problem

    I came across something used in another proof and I wanted to prove this simple fact for myself, although when I tried to pin is down with epsilons and the like, I found that I kept getting stuck! Considering sequences in $\displaystyle \mathbb{R}$. I would like to show that if $\displaystyle \frac{x_{n+1}}{x_n}\rightarrow 0$ then $\displaystyle x_n \rightarrow 0$.

    I would only like a hint with regards to how to do it, as I want to do it myself!

    If I could prove a slightly weaker conclusion, that $\displaystyle \{x_n\}$ converges, then I can use the fact that convergent implies bounded and then I'd have something like

    $\displaystyle |x_{n+1}| < \epsilon |x_n|$

    for suitably large n, then I could use the bound on $\displaystyle x_n$ to force $\displaystyle x_{n+1}$ to zero and therefor $\displaystyle \{x_n\} \rightarrow 0$.

    Am I over thinking this? Is there a simple proof? Some clever trick I have not seen?

    Any help would be much appreciated, thanks
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  2. #2
    MHF Contributor

    Aug 2006
    Is clear to you that from the given we have $\displaystyle \left| {\dfrac{{x_{n + 1} }}{{x_n }}} \right| \to 0?$
    Therefore, we know that the series $\displaystyle \sum\limits_n {x_n } $ converges.
    That implies $\displaystyle \left( {x_n } \right) \to 0$.
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