1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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

    User Avatar
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook