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Confusion regarding delta definition of limit

  1. Mar 12, 2013 #1
    I don't quite get the significance of the delta limit definition,

    if n>N and |sn−s|< ϵ , why does the limit converges

    does this simply means that there exist a number ε such that if n is great enough it will be greater than s by ε?

    But this doesn't make sense, because s is the value the sequence converges to, so sncannot be greater than s

    what confuses me even more is that (-1)n is a diverging sequence but
    it satisfies |sn-s| < ε for ε>1

    Can someone clarify this concept, thank you
     
  2. jcsd
  3. Mar 12, 2013 #2

    jbunniii

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    No, first of all you need to state the full definition: for EVERY ##\epsilon > 0## (no matter how small), there exists some ##N## such that ##|s_n - s| < \epsilon## for every ##n > N##. In general, choosing a smaller ##\epsilon## means that the required ##N## will be larger.

    Note that ##|s_n - s| < \epsilon## means exactly that ##-\epsilon < s_n - s < \epsilon## or equivalently ##s - \epsilon < s_n < s + \epsilon##. So convergence means that no matter how small an ##\epsilon## I choose, it's possible to find an ##N## such that ##s_n## will be contained within the interval ##(s-\epsilon, s+\epsilon)## for all ##n > N##.

    You didn't mention what ##s## is in this case. But note that it's not enough for the inequality to be true for ##\epsilon > 1##. It has to be possible to make it true for any ##\epsilon > 0##, no matter how small. To see that this is impossible, take ##\epsilon = 1/2## for example, and show that there can't be any ##s## that works.
     
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