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Converge of 1/ln(x)

  1. Feb 22, 2008 #1
    If derivative of 1/ln(x), which is -1/(x*ln(x)^2), converges
    why then 1/ln(x) does not converge?

    According to some theorem that I learned, differentiating does not change the radius of convergence and hence neither its convergence or divergence.

    Thanks.
     
  2. jcsd
  3. Feb 22, 2008 #2

    Gib Z

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    What are you talking about? Are you talking about a series, a sequence, integral, I have no idea. Also, it would help if you actually stated the theorem.
     
  4. Feb 22, 2008 #3

    VietDao29

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    Nah, there's no such thing like convergence of a function. A function is either defined, or undefined at some value.

    Seeing that you mention radius of convergence in your post, are you talking about power series?
     
  5. Feb 23, 2008 #4
    Yes, I was talking about power series. I was referring to The Term-by-Term Integration Theorem, .... (and other related ones_)
    I been so much into this stuff that I forgot that there's a summation sign that goes in the front.

    Thanks.
     
  6. Feb 23, 2008 #5

    VietDao29

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    =.=" Still not really sure what's troubling you. Is that the series concepts which confuse you, or you are not sure about convergent, and divergent tests, or.. what?

    The first post doesn't make much sense to me, though. :(

    If you are not sure about what troubles you, either. Then I suggest you spend some times, re-reading the whole chapter on series from the beginning thoroughly. Series is a pretty hard concept to grasp.

    When everything gets a bit clearer, and you're pretty sure which parts confuse you. You can post it here, and we may help. :)
     
  7. Feb 23, 2008 #6

    HallsofIvy

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    ?? The Taylor series for both 1/ln(x) and -1/(x ln(x)^2), about x= 1, converge with radius of convergence 1.
     
  8. Feb 23, 2008 #7
    How about:

    (1)sum (from n=2 to inf): 1/ln(n) ... diverges because 1/ln(n) >1/n and by comparison test this diverges
    (2)and sum (from n=2 to inf) = 1/(n*(ln n)^2) ... converges because it's integral limit is 1/ln(2) (so -1/(n*(ln(n))^2) also converges which is derivative of 1/ln(n))

    Am I missing something?
    And, I just remembered reading in the book that the theorem I mentioned above applies only to power series but 1/ln(n) is not a power series.
    I think I got it:
    (1) There is no radius of converges thing defined for non-power series but I was using it everywhere
    (2) theorems for power series (like R stays same) do not apply to non-power series.

    So, am I on right track now?
     
  9. Feb 23, 2008 #8
    Yes.
     
  10. Feb 23, 2008 #9

    HallsofIvy

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    You told us before you were talking about power series, and finally you tell us you really meant sum of ln(n)???
     
  11. Feb 23, 2008 #10
    somehow, I was thinking that there's no difference between sum of ln(n) and power series ><...
     
  12. Feb 23, 2008 #11
    Another Question:

    sum (k=0:inf) (-1)^(3k+1) / (2k+1) converges?

    I know sum (k=0:inf) (-1)^(k) / (2k+1) converges
    but sum (k=0:inf) (-1)^(2k) / (2k+1) not converges

    so how should I deal with such situations where I cannot say that the series is alternating, but it has same form?
    I just found that even*odd = even and odd*odd = odd

    so it does not matter whether it is:
    sum (k=0:inf) (-1)^(7k) / (2k+1) or sum (k=0:inf) (-1)^(9k+2) / (2k+1)
    the series is similar to sum (k=0:inf) (-1)^(k) / (2k+1) ??

    I think I pretty much got it but I got the answer while I was typing this question, so can you assure me that I have got the right answer?
     
  13. Feb 24, 2008 #12
    Try the root test:

    [tex]
    \begin{align*}
    \lim_{n \rightarrow \infty} a_n & < 1 \Rightarrow \sum a_n \mbox{ converges}\\
    & > 1 \Rightarrow \sum a_n \mbox{ diverges}\\
    & = 1 \Rightarrow \sum a_n \mbox{ might converge or diverge}
    \end{align*}
    [/tex]
     
  14. Feb 24, 2008 #13

    VietDao29

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    What is an that you are talking about? It doesn't seem no where near a root test to me. :(

    If a series converge, then its terms (an) tend to 0.

    (Notice that, this is a one way statement, the other way round is not correct, i.e, if its term tends to 0, the series can also be divergent, e.g [tex]\sum_{n = 1} ^ {\infty} \frac{1}{n}[/tex] diverge, whereas 1/n ~~> 0)

    The other equivalent statement is:

    If a series' terms don't tend to 0, then the series diverge.

    From here, what can you say about the series, is it convergent, or divergent?

    Those are all anternating series. ^^!

    You can check it. If 7k is odd, then 7(k + 1) is even; if 7k is even, then 7(k + 1) is odd, blah blah..

    Yup, sometimes, re-reading books proves to help quite a lot. See? :)
     
  15. Feb 24, 2008 #14
    Sorry. I meant
    [tex]
    \begin{align*}
    \lim_{n \rightarrow \infty} (a_n)^{1/n} & < 1 \Rightarrow \sum a_n \mbox{ converges}\\
    & > 1 \Rightarrow \sum a_n \mbox{ diverges}\\
    & = 1 \Rightarrow \sum a_n \mbox{ might converge or diverge}
    \end{align*}
    [/tex]
     
  16. Feb 24, 2008 #15
    Thanks a lot ;)

    but, I wonder if I really need a root test here (You can simply use AST)
     
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