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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.

    [tex]|a_n-a|<\epsilon[/tex]

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

    statdad

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    Homework Helper

    Are you sure

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


    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

    tiny-tim

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    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/(3-5)(3-2)
    -1/2 < 0

    and for n>1, it is monotonic.

    So if I write
    [tex]|a_n-a|=|\frac{2n-3}{3n-5}-\frac{2}{3}|=|\frac{3(2n-3)-2(3n-5)}{3(3n-5)}|=
    |\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?
     
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