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Diverging Series Proof; Basic

  1. Aug 16, 2014 #1
    Hi Guys,

    I am self teaching some basic analysis out of interest and I have a question on trying to prove a series diverges.

    Question; Prove that (-1)^n (1+1/n) diverges.

    My attempted approach was through contradiction:

    Assume that (-1)^n (1+1/n) converges. Then for some ε>0, there exists an N such that for all n>N,

    |(-1)^n (1+1/n) - L | < ε

    Then I broke down into two cases; n is odd, or n is even

    Suppose n is even, then we have | 1+1/n - L | < ε = 1 + 1/n < ε + L

    Now suppose that n is odd, then we have | -1 - 1/n - L | < ε = -1 - 1/n < ε + L

    Here is where I get stuck, I do not see where the contradiction is. Intuitively I know that the series jumps due to the (-1)^n and so alternately moves away from L, but I cannot show this.

    If you guys could point me to where I am going wrong I would greatly appreciate it.

    Thanks,
    ZZD
     
  2. jcsd
  3. Aug 16, 2014 #2
    You should probably break it down into the two cases where ##N## is odd or even. Is that what you meant?
     
  4. Aug 16, 2014 #3
    Thanks for your reply, I think I may have been mistaken. I broke down that n is odd or even, not N is odd or even.

    So doubling back and trying first that N is odd, I get the following:

    For the series to converge, I need some N, such that for all n>N, |(-1)^n (1+1/n) - L | < ε

    Now let N be odd. Then plugging in N is odd into the limit definition; |(-1)^N (1+1/N) - L | < ε

    is the same as | -1 - 1/N - L | < ε , which can be rewritten as 1 / N > -ε -1 -L

    EDITED (17:17 GMT)

    Ok I had another look, so from above, I would continue as saying that since 1 / N > -ε -1 -L, by moving stuff around, -1 / ε +1+L > N

    But if -1 / ε +1+L > N, this would imply that N is negative, since ε +1+L > 0, provided L is > 0. If N is negative it cannot be an element of the Natural Numbers. This then contradicts the definition of there being some N, which is a natural number, such that n>N, |(-1)^n (1+1/n) - L | < ε.

    Is this a bit better than before?
     
    Last edited: Aug 16, 2014
  5. Aug 16, 2014 #4

    Ray Vickson

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    What general facts do you know already about convergent sequences? Are you aware of the theorem that if a sequence converges (to limit L) then every infinite subsequence of it also converges to L? If you do know this, do you see how it applies to your current problem?
     
  6. Aug 16, 2014 #5
    Thanks Ray,

    My background knowledge is very limited. I literally asked someone to give their first year math notes, and began going through it yesterday. Prior to this I have never done any real proof based math in terms of exercises.

    In terms of what I know about sequences:

    1) A sequence of real numbers is a function f : N → R
    2) The sequence (an)n∈N is said to converge to L if for every real number ε > 0, there exists an N ∈ N (possibly depending on ε) such that for all n > N, |an −L|<ε
    3) If there does not exist a number L such that lim an = L, then the sequence (an)n∈N is
    said to be divergent

    I am not aware of the infinite subsequence fact you have stated. I would assume I would show that this sequence has two infinite subsequences with different limits, and then show that the limits are not equal, therefore implying the sequence itself does not satisfy the fact that all subsequences converge to a single limit. The two limits would arise from the fact that there is a (-1)^n i would assume, although not sure. Apologies if this is very basic.
     
  7. Aug 16, 2014 #6

    Ray Vickson

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    Basically, you've got it: the two sequences (for even n and for odd n) have different limits, so the entire sequence cannot have a limit. For large even n the terms are near +1 and for large odd n they are near -1. You don't even need to know the cited theorem to see why this is all you need, after some cleaning up.

    I think that a common source of difficulty for many students is that they insist on formalizing limit arguments by "epsilon-delta" type arguments right away, without maybe understanding intuitively what these are all about. IMHOP it is better to first understand in plain English and conceptually what a limit means, then translate the results into an epsilon-delta type argument at the end. In this case, we want to know: for larger and larger n, do all the members of the sequence come nearer and nearer to some common value?
     
  8. Aug 16, 2014 #7
    Thanks a lot. I will give it a shot:

    Suppose that N exists, and that there is some Limit, L, that all the subsequences converge to.

    First the even case: there is some very large even n>N, which is a natural number, such that
    |(-1)^n (1+1/n) - L1 |< ε , equivalently, 1 + 1/n - L1 < ε

    Odd case: suppose there is some very large odd n>N, which is a natural number, such that
    |(-1)^n (1+1/n) - L2 |< ε, equivalently, -1 - 1/n - L2 < ε

    It follows that since ε is fixed, and that 1/n in both cases is close to zero, roughly speaking 1-L1<ε and -1-L2<ε.

    Hence L1 and L2 cannot be equal, since each limit is close to 1 and -1 respectively.

    Since L1 and L2 are not equal, the series is not convergent, and therefore divergent.
     
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