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Use definition of limit

  1. Nov 8, 2004 #1
    Hi all, my task is to check, whether the given sequence has a limit and if yes, count it. We have to do it using the definition of limit.

    So I have eg. this sequence:

    (-1)^n \left( \frac{1}{10} - \frac{1}{n} \right)

    I know how the definition is, but I don't know how to use it for the purpose wanted. I just wrote

    \left| A - (-1)^n \left( \frac{1}{10} - \frac{1}{n} \right) \right| < \epsilon , \forall \epsilon > 0

    But how to prove that the sequence has or has not limit? Should I just try to prove existence of the limit, or, on the contrary, should I try to prove that the limit doesn't exist? What is the general recommended method, when we have to prove it from definition of limit?

    Thank you all for any answer.
  2. jcsd
  3. Nov 8, 2004 #2


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    There is no general solution for finding the limits of sequences. In fact, there are a goodly number of outstanding conjectures that the limit of a particular sequence is some value. In this particular case, it should be possible to show that the sequence does not tend to a limit.

    Hint: If the sequence [tex]\{s_n\}[/tex] has a limit converges, then the sequence[tex]\script{S}_n=s_n-s_{n+1}[/tex] will tend to zero.
    Last edited: Nov 8, 2004
  4. Nov 8, 2004 #3
    Well, so I always have to decide, whether I will try to prove the existence of the limit, or the opposite?

    Is it a general theorem? If not, when can I use it? I'm asking because I think we didn't have this theorem so far...
  5. Nov 8, 2004 #4


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    If you replace 'has a limit' with 'converges' (which is what I should have written it initially), then it's generally true.

    Consider that if [tex]|s_n-A| < \epsilon[/tex] then [tex]s_n \in (A-\epsilon,A+\epsilon)[/tex]. Now, if
    the sequence converges, then given any [tex]\epsilon > 0[/tex] it's possible to find [tex]N_\epsilon[/tex] so that [tex]n>N_\epsilon \Rightarrow |s_n - A| < \frac{\epsilon}{2}[/tex]
    [tex]n > N_\epsilon \Rightarrow s_n \in (A-\frac{\epsilon}{2},A+{\epsilon)}{2}[/tex]
    This means that the 'tail' of the sequence is contained in an interval with a length (diameter really) of [tex]\epsilon[/tex].
    So, cleary for
    [tex]n_1,n_2 > N_\epsilon [/tex] we have [tex]|s_{n_1}-s_{n_2}| < \epsilon[/tex]
    [tex]|s_{n_1}-s_{n_2}| = |s_{n_1}-s_{n_2}-0| < \epsilon[/tex]
    So the differences go to zero if the sequence coverges.

    The notion of the difference between elements in the tail of a squence going to zero if the sequence coverges is important in real analysis:
    Last edited: Nov 8, 2004
  6. Nov 9, 2004 #5

    matt grime

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    If a sequence converges, the limit is unique and every subsequence also converges to that limit.

    So, if you're sequence has two subsequences tending to distinct limits then it cannot itself converge.

    Can you see how that helps in your example above?
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