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Prove the convergence of a limit

  1. Dec 17, 2012 #1
    1. The problem statement, all variables and given/known data
    it should be all right this time, but could you please check my solution?

    prove the convergence and find the limit of the following sequence:
    ##a_1>0##
    ##a_{n+1}= 6 \frac{1+a_n}{7+a_n}##
    with ## n \in \mathbb{N}^*##

    3. The attempt at a solution
    the sequence is increasing, and its two possible limits are the solution of the associated equation: 2; -3.
    However, I have to exclude -3, because being ##a_1>0## and the sequence increasing, it can't converge to a negative number.
    Saying it converges is equal to saying that:
    for every ##\epsilon<0## there exists a n* s.t. for every n≥n* we have:
    ##|a_n-L|<\epsilon##
    being L=2 in this case and ##a_{n+1}>a_n>a_{n*}## we have
    ##\frac{6+6a_n}{7+a_n}-2|<\epsilon##
    and
    ## |\frac{6- 14 -2a_n+6a_n}{7+a_n}|<\epsilon##
    which holds for every
    ## a_n> \frac{-7\epsilon-8}{4-\epsilon}##
     
  2. jcsd
  3. Dec 17, 2012 #2

    SammyS

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    ##\displaystyle \quad\quad \quad\quad \left|\frac{6+6a_n}{7+a_n}-2\right|<\epsilon##
    You have a few typos in your post. I cleaned some up and enlarged some fractions.



    ##\displaystyle \frac{-7\epsilon-8}{4-\epsilon}\ \ ## is negative for ε < 4 .

    If a1 > 2 , then the sequence is decreasing.
     
  4. Dec 18, 2012 #3
    great, thanks!
    would it be better then, if i wrote:
    which holds for
    ##\displaystyle \frac{-7\epsilon-8}{4-\epsilon}>0##
    with
    ##\epsilon>4##
    ?
     
  5. Dec 18, 2012 #4

    SammyS

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    No. Not at all.

    You need to show that [itex]\displaystyle \ \left|\frac{6+6a_n}{7+a_n}-2\right|<\epsilon\ \ [/itex] for all ε > 0. The crucial situation is when ε is very small.
     
  6. Dec 18, 2012 #5
    should i put it like that:
    ##|\frac{7\epsilon+8}{4-\epsilon}| > a_n## which for ##\epsilon## going to 0, goes to 2? --> ##a_n<2##
     
  7. Dec 18, 2012 #6

    SammyS

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    What you need to show is that for all n > n*, [itex]\displaystyle |a_n-2|<\varepsilon\ .[/itex]

    If you can find a way to define an in terms of n and a1, then the ε-n* proof described above is appropriate.

    I have worn out WolramAlpha, messing around with this problem.

    I haven't worked with recursively defined sequences lately... but here's an idea.

    If the sequence an is monotonically increasing, which it is if 0 < a1 < 2, and if the sequence has an upper bound of 2, then all you need to show is that for ε > 0, there is a n>0 for which ak > 2-ε .

    Now that I consider that, that's not easy to do either.
     
  8. Dec 19, 2012 #7

    pasmith

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    Acutally if [itex]a_1 > 2[/itex] then the sequence is monotonic decreasing. What is true is that if [itex]a_n > 0[/itex] then [itex]a_{n+1} > 0[/itex], so if [itex]a_1 > 0[/itex] then certainly the sequence cannot converge to -3.

    With sequences defined by recurrence relations, it's best to note that if [itex]|a_n - L|[/itex] is monotonic decreasing, then it is bounded below (by 0, as absolute values always are) and so [itex]|a_n - L|[/itex] must converge to a limit. Hence [itex]a_n[/itex] converges, and given how [itex]L[/itex] was determined [itex]a_n[/itex] must converge to [itex]L[/itex].

    The condition for [itex]|a_n - 2|[/itex] to be monotonic decreasing is that [itex]|a_{n+1} - 2| < |a_n - 2|[/itex] for all n. Now
    [tex]
    \left|a_{n+1} - 2\right|
    = \left|6\frac{a_n+1}{a_n+7} - 2\right|
    = \left|\frac{4a_n - 8}{a_n + 7}\right| = \frac{4}{|a_n + 7|}|a_n - 2|
    [/tex]
    which means that [itex]|a_{n+1} - 2| < |a_n - 2|[/itex] if [itex]|a_n + 7| > 4[/itex]. This holds, in particular, if [itex]a_n > 0[/itex]. Since [itex]a_1 > 0[/itex] and if [itex]a_n > 0[/itex] then [itex]a_{n+1} > 0[/itex], [itex]|a_{n+1} - 2| < |a_n - 2|[/itex] for all n. Hence [itex]a_n[/itex] tends to a limit, which in the circumstances must be 2.
     
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