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Homework Help: Series: converges or diverges help

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Determine whether the series [PLAIN]http://img265.imageshack.us/img265/69/44751948.jpg [Broken] converges or diverges and find its sum S if it does converge.

    2. Relevant equations

    3. The attempt at a solution
    my textbook has lots of weird examples but none of them help me understand how to solve this one..
    if i plug in 1 then 2 then 3.... i get
    ln(1.5) + ln(1) + ln (7/8) +...+ln((2n+1)/(3n-1))

    so a = ln(1.5)
    r = ?

    then i use the formula: a / (a-r)
    right? now how do i find r?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 6, 2010 #2

    What can you say about [itex]\frac{2n+1}{3n-1}[/itex] as [itex]n\to\infty[/itex]? And then about [itex]\ln\left(\frac{2n+1}{3n-1}\right)[/itex]? Now considering adding a bunch of terms that eventually behave like [itex]\ln\left(\frac{2n+1}{3n-1}\right)[/itex]
    Last edited by a moderator: May 4, 2017
  4. Apr 6, 2010 #3


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    Homework Helper

    worth checking what the nth term does for large n
  5. Apr 6, 2010 #4
    [itex]\frac{2n+1}{3n-1}[/itex] as [itex]n\to\infty[/itex] will be 0
    and for [itex]\ln\left(\frac{2n+1}{3n-1}\right)[/itex] it will be ln(0) which is impossible..

    but if plugging numbers like 10, 100, 1000..... into [itex]\ln\left(\frac{2n+1}{3n-1}\right)[/itex] the number becomes smaller and smaller all the way to -[itex]\infty[/itex]
    so does it mean that the answer is 'Diverges'?
  6. Apr 6, 2010 #5

    Char. Limit

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    Gold Member

    As n goes to infinity, [itex]\frac{2n+1}{3n-1}[/itex] does not go to zero.
  7. Apr 7, 2010 #6
    oh my bad.. its 1
    ln(1) = 0

    so the answer is Diverges, S=0. ?
  8. Apr 7, 2010 #7


    Staff: Mentor

    Wow! Two mistakes in only three lines!

    (2n + 1)/(3n - 1) doesn't approach 1, so ln( (2n + 1)/(3n - 1)) doesn't approach 0.

    Furthermore, in a series [itex]\sum a_n[/itex], if lim an = 0, you can't conclude much of anything. Consider [itex]\sum 1/n[/itex] and [itex]\sum 1/n^2[/itex]. In both series an --> 0, but the first series diverges and the second series converges.
  9. Apr 7, 2010 #8
    Sorry. This is incorrect.

    I can not speak too much about series convergence/divergence, but I can talk about a sequence:

    What happens to [tex]\frac{2n+1}{3n-1}[/tex] as n ---> infinity?

    You can rig this so that the 1's cancel and you are then left with [tex]\frac{2n}{3n}[/tex] which obviously then cancels down to [tex]\frac{2}{3}[/tex] (i.e. the limit of the sequence).

    Now as you are actually dealing with a natural log, the limit of the sequence in actual fact is [tex]ln(\frac{2}{3})[/tex]

    If you wished to prove this you could do the following for all n > N: [tex]|ln (\frac{2n+1}{3n-1}) - ln(\frac{2}{3})|<\epsilon [/tex]

    Also, note that this sum is not a geometric progression: you find r as such: [tex]r = \frac{t_{n}}{t_{n-1}} [/tex] where [tex] t_{n} [/tex] is the nth term of the series. Take note of the second term of the sequence, it equals zero, and division by zero is undefined. Also, r is the common ratio. So the series is, by definition, not geometric in nature.

    I am not absolutely sure of this, but I believe the series in fact diverges to negative infinity, because of the simplification of the natural log as per above. You have an infinite number of ln(2/3) to multiply out which is equal (in a very loose sense) to negative infinity. I may be wrong however.
    Last edited: Apr 7, 2010
  10. Apr 7, 2010 #9
    The series diverges, but not due the reasons you pointed out (which were not only incorrect).

    Think about what happens if the n-th term in your sum does not tend to 0. Why does the following series diverge?

    [tex]\sum_{n=1}^\infty k = k + k + k + k + k + \dotsm[/tex]

    where [itex]k\not=0[/itex]. Think about how this relates to the series in the original problem.
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