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Limit of a sequence

  1. Mar 22, 2006 #1
    hi, can someone please look at this. it doesnt look right to me. thanks

    i have to find the limit:

    a(1)= 2 a(n+1)=1/(3-an)

    a1=2, a2=1, a3= .5, a4= .4, a5= .3846, a6=.382385... a9=.3819742, a10= .3819672

    therefore the limit would be approximately .38197 right?
     
  2. jcsd
  3. Mar 22, 2006 #2

    Curious3141

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    Well, I think the question wants you to find it exactly.

    Are you familiar with continued fractions ?
     
  4. Mar 22, 2006 #3

    Curious3141

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    OK, let me try to get you started.

    At the limit (if it exists), the successive values shouldn't change.

    That is,

    [tex]lim_{n \rightarrow \infty} a_{n+1} = a_n[/tex]

    In other words,

    [tex]lim_{n \rightarrow \infty} \frac{1}{3 - a_n}= a_n[/tex]

    Can you solve that equation ? There will be 2 solutions, only one of which is admissible. But to see why only that one should be accepted you have to do some further work. Do this first, then we'll take it from there.
     
  5. Mar 22, 2006 #4
    Rather

    [tex]\lim_{n \rightarrow \infty} a_{n+1} = \lim_{n \rightarrow \infty} a_n[/tex]

    and similarly for the other identity.
     
  6. Mar 22, 2006 #5

    Curious3141

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    What's the difference from what I posted ?

    Maybe the LaTex wasn't clear (I hate Tex), but I intended the limit to apply to both LHS and RHS. I don't see anything wrong with my notation, do you ?
     
    Last edited: Mar 22, 2006
  7. Mar 22, 2006 #6
    you guys are great!

    thanks
    :!!)
     
  8. Mar 22, 2006 #7
    Curious, your notation is saying that [itex] a_n [/itex] is the solution to [tex] \lim_{n \rightarrow \infty} a_{n+1}[/tex]. Muzza's notation is saying that the two limits are equivalent.
     
  9. Mar 22, 2006 #8

    HallsofIvy

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    It was really your statement "At the limit (if it exists), the successive values shouldn't change." that is not very precise. At the limit, there are no "succesive values"!

    Of course, it is true that {an} and {an+1} are just renumberings of the same sequence and so must have the same limit. If the sequence has a limit, call it "a" and then
    [tex]a= lim a_{n+1}= lim \frac{1}{3- a_n}= \frac{1}{3- lim a_n}= \frac{1}{3- a}[/tex]
     
  10. Mar 22, 2006 #9

    Curious3141

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    The spacing is off. The limit applies to both terms, more like :

    [tex]\lim_{n \rightarrow \infty} (a_{n+1} = a_n)[/tex]
     
  11. Mar 22, 2006 #10

    Curious3141

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    But successive values are in fact what we're talking about, whether we want to overtly admit it or not. What we're (both) saying is in effect that with higher and higher ordinal numbers, the difference between successive terms become smaller and smaller. At the limit, the difference between successive terms vanishes !

    OK, it looks "better" to call the limit a but it's essentially the same thing, IMHO. The essence of the whole thing is that :

    [tex]\lim_{n \rightarrow \infty} (n+1 = n)[/tex]

    is it not ? Or if you'd prefer to break it up into two limits,

    [tex]\lim_{n \rightarrow \infty} (n+1) = \lim_{n \rightarrow \infty} n[/tex]
     
  12. Mar 22, 2006 #11

    shmoe

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    This is not notation I've ever seen used, anywhere. In my opinion it's not a terribly good notation either. It looks like you're somehow distributing the limit over the = sign?

    Just write the limit on both sides of the equation, and end all misunderstandings.
     
  13. Mar 22, 2006 #12

    Curious3141

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    OK, if it's not standard notation, then I stand corrected. :smile:
     
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