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Limits superior and inferior

  1. Feb 2, 2013 #1
    (Hello everyone!)
    1. The problem statement, all variables and given/known data
    Given that [tex]\limsup_{n \rightarrow \infty}(\frac{1}{x_n})\cdot \limsup_{n \rightarrow \infty}(x_n)=1[/tex]
    Show that x_n converges.

    2. Relevant equations
    Recalling that:
    [tex] x_n \text{ converges } \iff \liminf(x_n)=\limsup(x_n)[/tex]
    3. The attempt at a solution
    Started with this:
    Firstly, let's designate the limsup of (x_n) as L;
    Therefore, according to the stipulated conditions, limsup(1/x_n) = 1/L;
    [tex]1. \forall \epsilon >0 \; \exists N \mid \forall n>N \mid x_n \leq L+\epsilon[/tex]
    However, for the limit inferior:
    [tex]2. \forall \epsilon >0 \; \exists N^{*} \mid \forall n>N^{*} \mid x_n \geq L-\epsilon[/tex]
    Here arises my first question: can I unite both Ns, by finding the *maximum* of both, i.e. N'=Max{N, N*}?(Even though, I am not sure that in fact, higher orders of x_n, would find themselves in the inferior limit region).
    And if not, how do I proceed?
    (I tried inverting ineq. 2), to arrive at a value for limsup(1/x_n), but, to no avail.
    Any help will be much appreciated,
    (And I'm sorry if I bungled some of the definitions here, it's been a while since I had last done this :)),
    Most beholden,
    And grateful as always,
    Last edited: Feb 2, 2013
  2. jcsd
  3. Feb 2, 2013 #2
    What do you think of this(is this valid?); some limited progress:
    1. \limsup(x_n) = L \Rightarrow \forall \epsilon >0 \mid \exists N_1 \mid \forall n>N_1 \Rightarrow
    x_n < L+\epsilon \\
    2. \limsup(\frac{1}{x_n}) = \frac{1}{L} \mid \forall \epsilon >0 \Rightarrow \exists N_2 \mid \forall n>N_2 \Rightarrow \frac{1}{x_n} < \frac{1}{L}+\epsilon \\
    3. N^{\prime} = \max(N_1,N_2) \\
    4. \text{Rearranging 2. : if } \frac{1}{x_n} < \frac{1}{L}+\epsilon \Rightarrow x_n > \frac{L}{1+\epsilon L} \\
    5. \text{via 1 & 4(assuming N' >= N1, N2(3.)): } L+\epsilon > x_n > \frac{L}{1+\epsilon L} \\
    6. \text{Setting epsilon: } \epsilon = \frac{a}{n} \mid a \in \mathbf{R} \\
    7. \text{Using the "Squeeze Theorem" :}h(n) = L+\epsilon(n) \rightarrow L \\
    g(n) = \frac{L}{1+\epsilon(n) L} \rightarrow L \\
    And so you obtain that x_n converges, and that in fact, as we would expect, to L.
    The problem is, can I use the Squeeze theorem, even though I am not certain that x_n, in fact, converges? Or is it simplify sufficient to point out, that since I have already found "N" -- that very N could be translated to a direct proof of the convergence of x_n, via the epsilon/N notation?(i.e. by definition)?
    Any thoughts?
    Again, very thankful for your attention,
    (All limits tend to infinity, as stated in the OP)
    Last edited: Feb 2, 2013
  4. Feb 2, 2013 #3
    Haven't figured the problem out yet, but here are some thoughts:
    - Where in your proof do you use the given statement?
    - In step 6, can you just set epsilon, isn't epsilon supposed to be random >0?
  5. Feb 2, 2013 #4
    Thanks for your reply M. Planck!
    1. I used the initial conditions, when setting the Limsups; That is, I had first declared limsup(x_n) to be some arbitrary(but real) L; then, using the provided relation, I educed limsup(1/x_n) to be 1/L.
    2. Setting epsilon thus still provides me with randomness(since "a" could be any real number). However, that was done merely to bound x_n. Since the derived inequality for x_n(5.) applies to all epsilons(greater than 0), then surely it would still hold for diminishing values of epsilon as well.
    This gave me the necessary circumscription of x_n, to use the "squeeze theorem"; of the verity of whose usage, however, I am still uncertain.
    What do you think, though?
    Once again,
    Very grateful,
  6. Feb 2, 2013 #5
  7. Feb 2, 2013 #6
    I am not quite so sure about that;
    You see, as far as I understand it, Limsup, and Liminf don't follow the arithmetical rules of (finite) Limits.
    In other words, I am not at all aware that Limsup(1/x_n) = 1/Limsup(x_n) = ...
    I believe that's part of the reason why that specific condition was given.
    However, if you believe this to be the case, I would very much welcome a link to a proof of that particular rule. It could very useful in future.
    Also, why should N1 equal N2? could you elaborate on that? in any case, I thought taking the maximum eliminated any need to specify them, or find a particular congruence thereof.
    Looking forward to your replies,
  8. Feb 2, 2013 #7
    Never mind, what I said is not true.

    I think your proof is correct. Now I see how you used the condition.
  9. Feb 2, 2013 #8


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    Yes, if x_n can be squeezed between two sequences that both converge to the same limit, then that forces x_n to converge, to the same limit.

    I think your proof is mostly fine, but you are implicitly assuming (e.g. in step 4) that x_n > 0 for all n. I didn't see that stated as a hypothesis. Isn't the theorem still true if some or all of the x_n's are negative? (I think we can presume that they all have to be nonzero, so that 1/x_n is defined.)
  10. Feb 2, 2013 #9
    M. Planck, thanks again for all your help; I hope my approach really does check out :)...
    Great comment, thanks. As I said, I haven't tinkered with these types of problems for quite a while, and, I completely forgot to generalise the formulae for x_n < 0 (obviously, x_n must be non-zero).
    The point is, I'm not really sure how this is supposed to work.
    I know from the definition of Limsup that x_n(with n>N) lies to the left of L+eps. But, it is not at all certain, that ALL elements of x_n (n>N) are in fact larger than L-eps.
    Therefore, taking the absolute value of x_n is impermissible here.(i.e. |x_n-L|<eps --> which is incidentally the definition for a finite, complete limit).
    I also tried to keep L(when manipulating it) as arbitrary and nondescript as possible(save for the fact of course, that it's non-zero). I thought that sufficiently covered it.
    Wouldn't negative x_n-s simply invert the direction of the squeezing inequality? meaning:
    h(n)>x_n>g(n) in that case?
    I am also not so sure what to do, in that case, about "ill-behaved", pathological sets(even though I think the fact that limsup(x_n)*limsup(1/x_n) = 1 > 0 pretty much excludes those sets).
    How do you propose to generalise this?
    I remain extremely grateful for all your help!
    Last edited: Feb 2, 2013
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