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!

Homework Help: Progression, is monotonic?

  1. Oct 4, 2008 #1
    1. The problem statement, all variables and given/known data

    I got one progression [tex]\frac{2n-3}{3n-5}[/tex].

    Is this monotonic and convergent?

    2. Relevant equations

    3. The attempt at a solution

    I tried an-an+1=1/(3n-5)(3n-2)>0

    But for n=1 and n=2, we got 1/2 and 1, so I think that this string is not monotonic, right?

    I think it is convergent because it got a=2/3.


    Am I right?
  2. jcsd
  3. Oct 4, 2008 #2


    User Avatar
    Homework Helper

    Are you sure

    a_n - a_{n+1} = \frac 1 {(3n-5)(3n-2)}

    is [tex] > 0 [/tex] for every [tex] n = 1, 2, 3, \dots [/tex]? :smile: Compare the result you
    get from the formula above to the actual value of [tex] a_1 - a_2 [/tex].
    You are on the correct track for proving convergence.
  4. Oct 4, 2008 #3


    User Avatar
    Science Advisor
    Homework Helper

    Hi Physicsissuef! :smile:

    It would be much easier if you you rewrote it in the form 2/3 + A/(3n - 5). :wink:

    If it's monotonic except for the first one or two, then you should say so.

    It converges monotonically after a certain point. :smile:
  5. Oct 5, 2008 #4
    Yes, I am sure that it is correct, but as you can see for n=1 (i.e a1,a2), it is

    -1/2 < 0

    and for n>1, it is monotonic.

    So if I write
    |\frac{6n-9-6n+10}{3(3n-5}|=\frac{1}{3(3n-5)} < \epsilon[/tex]

    So it is convergent, probably for n>1, because if n=1, [tex]1/3(3-5)=-1/6 < \epsilon[/tex]

    Because of the fact that [tex]\epsilon[/tex] can't be negative i.e [tex]\epsilon > 0[/tex], it is convergent for n>1, right?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook