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Proving Divergence: is this a sufficient proof?

  1. Jun 27, 2011 #1

    Say you want to use a proof by contradiction to prove that a sequence diverges. So you assume that x(n)-----> L , and try to find a real number, call it M, such that |x(n) - L| can never get smaller than M, thus arriving at a contradiction.

    My question is: can M be of the form that it includes previous terms in the sequence? For example, is it sufficient to say:

    |x(n) - L| >= |(4 - L)/4x(n-1)| = M, for all natural n.

    Notice that the number I'm trying to use as a lower bound for the distance between x(n) and the supposed limit includes the previous term in the sequence as an inverse factor.

    I'm having trouble wrapping my head around this because my intuition tells me that this is insufficient, since you could just go farther out in the sequence to eventually get smaller than the original term's M. However, every term in the sequence will have it's own M which depends on the previous term in the sequence. I'm having trouble determining what this means.

    Any help please?
  2. jcsd
  3. Jun 27, 2011 #2
    I apologize in advance that I did not read the rest of your question. You have a misunderstanding in what you need to prove. I think once you work through the following example, the nature of convergence will be more clear.

    Consider the sequence 1/2, 47, 1/4, 47, 1/8, 47, 1/16, 47, ...

    where the terms alternate between the powers of 1/2 and the number 47.

    I claim you can not find any number M such that |x(n) - 0| never gets smaller than M. In fact given any [itex]\epsilon > 0[/itex], there are infinitely many n such that |x(n) - 0| < [itex]\epsilon[/itex].

    Yet, this sequence does not converge to zero.

    Why not?
  4. Jun 27, 2011 #3
    Clearly your sequence does not converge to zero because there does not exist N such for all n>=N abs(x(n))< epsilon

    I suppose I should've also stated that my sequence is monotone.

    I'm trying to find a number M where i can show |x(n) - L| NEVER gets smaller than M. Then, since there are NO terms within L distance to M, I will have proved divergence.

    I've found an expression for such an M, however for each x(n), my expression for M depends on the previous x(n-1) term. So my M is constantly changing with each term, and I'm trying to figure out if that is still logically okay?
  5. Jun 29, 2011 #4
  6. Jun 29, 2011 #5
    That makes a huge difference!

    This is all kind of vague, perhaps you could say what the sequence is and then people can provide specific suggestions.

    If you have a monotone sequence of reals, then if it is bounded it converges; and if it's unbounded it diverges to +/- infinity.

    So I wonder if taking some arbitrary L and then doing some calculation to show that your sequence never gets close to L, is possibly not the most efficient way to go about it.

    Your sequence either goes to +/- infinity or converges. Try working back from that.

    In other words if it diverges, it diverges BIG. You don't need some delicate calculation to show that it stays away from some number L. If it diverges, it gets HUGELY far from any finite number.

    See if any of this helps ...

    Last edited: Jun 29, 2011
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