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Divergence of 1/k

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data

    If lim(k>inf) 1/k, goes to 0, why does it diverge?

    2. Relevant equations

    Divergent series test

    3. The attempt at a solution

    i dont understand why 1/k (harmonic series) diverges, when according to the divergent series test, it should converge to 0.
     
  2. jcsd
  3. Apr 16, 2015 #2

    WWGD

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    What is the "divergence series test"? Maybe I(we) know it by a different name? But the n-th term going of the series going to 0 is definitely necessary, but not sufficient for convergence of the series.
     
  4. Apr 16, 2015 #3
    Yes that is the same test. So is it just a confirmation for divergence, but inconclusive at L = 0?
     
  5. Apr 16, 2015 #4

    LCKurtz

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    "It" doesn't diverge. ##\frac 1 k \to 0##.

    Be careful about distinguishing ##\frac 1 k## with the series ##\sum \frac 1 k##. The series and the sequence of its terms are different things.
     
  6. Apr 16, 2015 #5
    I'm sorry, I should've paid attention to the wording of my question. It should have been, why doesn't the series ##\sum \frac 1 k## converge? I was under the impression that if the limit of the series goes to 0, then the series converges. Yet if it doesn't it diverges. If ##\lim_{k\to\infty} \sum\limits_{k=1}^{\infty} \frac 1 k## = 0, then it converges? I don't really understand.
     
  7. Apr 17, 2015 #6

    LCKurtz

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    That doesn't make any sense. A series either converges, in which case it has a sum, or it diverges. There is no "limit of a series".

    Again, that makes no sense. The ##k## in the summation is a dummy variable. There is no ##k## to take the limit of.

    You are still confusing the limit of the ##k##th term with something else. The relevant theorem is : If the series ##\sum a_k## converges, then ##a_k\to 0##. The contrapositive statement is that if ##a_k## doesn't converge to ##0##, the series diverges, which is why it is sometimes called a divergence test.

    What the theorem doesn't say is that if ##a_k\to 0## then ##\sum a_k## converges. That is the converse of the theorem and it isn't true.
     
  8. Apr 17, 2015 #7

    Mark44

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    Your limit makes no sense, since in the sum, k takes on values 1, 2, 3, ... and so on.

    Take another look at the wording of the theorem you are using -- the mistake you are making is one that many students make. For a series ##\sum_{k = 1}^{\infty} a_k##, the Nth Term Test for Divergence, as it is usually called, says that if lim ak ≠ 0, the series diverges. The mistake that many new students make is in thinking that if if lim ak = 0, the series converges. THIS IS NOT TRUE, and has no connection with the Nth Term Test for Divergence. All it does is say whether a series diverges or not. It DOES NOT tell you that a given series converges.

    Edit: LCKurtz beat me to it. The form of this theorem he cited is equivalent to what I wrote.
     
  9. Apr 17, 2015 #8

    Ray Vickson

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    You have seriously misunderstood an important fact about infinite series: for convergence of ##\sum_{n=1}^{\infty} a_n## it is necessary to have ##a_n \to 0## as ##n \to \infty##. However, that is not sufficient; in fact, the series ##\sum 1/n## is a good counterexample.

    There are lots of proofs of divergence, but the easiest one is based on comparison with an integral. We have ##1/1 \geq 1/x## for ##1 \leq x \leq 2##, ##1/2 \geq 1/x## for ##2 \leq x \leq 3##, ##\ldots##. In general, ##1/k \geq 1/x## for ##k \leq x \leq k+1##. Thus, ##1/1 \geq \int_1^2 dx/x##, ##1/2 \geq \int_2^3 dx/x##, etc, so
    [tex] \sum_{n=1}^N \frac{1}{n} \geq \left( \int_1^2 + \int_2^3 + \cdots + \int_N^{N+1} \right) \frac{dx}{x} = \int_1^{N+1} \frac{dx}{x} = \ln(N+1)[/tex]
    This implies ##\sum_{n=1}^N 1/n \to \infty## as ##N \to \infty##, so the series diverges.
     
  10. Apr 17, 2015 #9
    Thank you, all of you. I finally understand. I guess that's the question I was trying to state all along when I said

    I just meant, "Does it prove the series diverges, but not necessarily that the series converges if L = 0"

    Thank you again for your time.
     
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