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Show that (-1)^n * (n/(n+1)) is divergent

  1. Sep 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Using a proof by cases, show that (-1)^n * (n/(n+1)) is divergent


    2. Relevant equations

    A sequence a_n is said to be convergent if
    For every real number ε > 0, there exists a natural number N such that for all n > N, |a_n − L| < ε.

    3. The attempt at a solution
    Tried to make n the next odd integer (2k+1) but struggled to go any further.
     
    Last edited: Sep 10, 2013
  2. jcsd
  3. Sep 10, 2013 #2

    UltrafastPED

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    Your expression is unclear (to me) ... is it supposed to be a summation? Is the n/(n+1) a factor or part of the exponent?

    At a guess you mean the sum -1/2 + 2/3 - 3/4 + 4/5 ... starting with n=1.

    Also I don't understand what you mean by "cases" ... what would the cases be?
     
  4. Sep 10, 2013 #3
    Sorry if I wasn't clear. The expression is a sequence starting from 1 through the Natural numbers. n/(n+1) is a factor not part of the exponent.
    The cases are
    L<0
    L=0
    L>0

    Where L is the limit of the sequence.
     
  5. Sep 10, 2013 #4

    UltrafastPED

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    A sequence is a list of numbers {1,2,3,...}; it may have a limit point, but not a limit.

    An infinite sum is defined to be the limit point of its sequence of partial sums; the infinite summation does not have a limit - it either has a value or is divergent.

    So I don't understand what L is supposed to represent - though I do understand the case structure.

    Please write out a few of the terms ... was my guess correct?
     
  6. Sep 10, 2013 #5
    Your guess is correct but it is not a sum, just a sequence. I had a look around and found a similar question albeit slightly simpler: https://www.physicsforums.com/showthread.php?t=491938
    I'm struggling to emulate the techniques used in that thread.
     
  7. Sep 10, 2013 #6

    UltrafastPED

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    The sequence is {-1/2, +2/3, -3/4, +4/5, ...} with an assumed limit point L.

    Proof that this sequence diverges by the method of contradiction in three cases: limit point L assumed to by +,0,-.

    Is that the problem?
     
  8. Sep 10, 2013 #7
    Yes that is the problem, sorry it took so long to get across.
     
  9. Sep 10, 2013 #8

    UltrafastPED

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  10. Sep 10, 2013 #9
    Last edited: Sep 10, 2013
  11. Sep 10, 2013 #10

    Mark44

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    The sequence as described in the OP is clear to me.
    There was nothing in the OP that would lead someone to think that this was a problem about a series.
    ???
    It is quite common to talk about the limit of a sequence. For example, with the sequence {an}, where an = ##\frac{n}{n + 1}, \lim_{n \to \infty}a_n = 1##.
     
  12. Sep 10, 2013 #11

    Mark44

    Staff: Mentor

    PhyzX, if you haven't already done so, look at the first five or six terms in your sequence. The terms jump around, due to the (-1)n factor.
     
  13. Sep 10, 2013 #12
    Hey, yeah I've noticed that. The sequence converges to both 1 and -1 if I'm not mistaken? However I am told to use the method of finding suitable epsilon and N in each case to produce counter examples. So I cant use subsequences and such to show that it converges twice.
    However I think I've got it now.

    After I substitute a value for n, say the first even integer greater than N, (n=2k), I'm left with something like | (2k)/(2k+1) - L | and since k is greater than 1, I can manipulate the terms to show that it is bigger than say ε=1/4
     
  14. Sep 10, 2013 #13

    Mark44

    Staff: Mentor

    No, that's not right. If a sequence converges, it can converge to only one value. It can't converge to both 1 and -1. Your sequence has a subsequence that converges to 1 and another that converges to -1, but the sequence itself doesn't converge.
    Keep in mind that when N is large, the terms in your sequence are bouncing around between numbers that are close to 1 and -1. When one term in your sequence is close to +1, the next term is going to be almost 2 units away, near -1. Similarly, when one term in the sequence is near -1, the next term will be again almost 2 units away, near +1.
     
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