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Homework Help: Limit of Sequence

  1. Nov 11, 2006 #1
    [tex]a_n = 2^n/(3^n + 1)[/tex]

    Find the limit as n -> infinity

    So I separated the fraction into [tex] (1/3)(2/3)^n [/tex]

    Then limit = 1/3 [tex] lim_(n->infinity) (2/3)^n. [/tex]

    So, my question is, how do you find the limit [tex] (2/3)^n. [/tex] Is analysis of functions simply enough, i.e. looking at how fast they increase?
     
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  3. Nov 11, 2006 #2

    quasar987

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    How did you transform a_n into [tex] (1/3)(2/3)^n [/tex]???

    Btw - is this for a real analysis class or some intro calculus class?
     
  4. Nov 11, 2006 #3
    Oops, sorry, I guess I didn't look hard enough at my post.

    It's really [tex] a_n = \frac{2^n}{3^{n+1}} [/tex]

    And it's Calculus II
     
  5. Nov 12, 2006 #4

    quasar987

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    Oh ok.

    Didn't you see the general result that if -1<a<1, then [itex]a^n\rightarrow 0[/itex]?

    If not, and if you've seen a little bit of series theory, you can prove it this way: You know that the geometric series [itex]\sum a^n[/itex] converges for -1<a<1 and diverges otherwise. And you've also seen the basic necessary criterion for a series to converge, namely that the general term [itex]a_n[/itex] of any converging series goes to zero in the limit n-->infty.

    So you can make use of these two fact by saying "I know that the geometric series of general term [itex]a_n=a^n[/itex] (where -1<a<1) converges, hence it must be that [itex]a_n=a^n\rightarrow 0[/itex] for -1<a<1."
     
    Last edited: Nov 12, 2006
  6. Nov 12, 2006 #5
    Ok, I get what you're saying...but why is -1<a<1 here?
     
  7. Nov 12, 2006 #6

    quasar987

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    I proved that for any real number a btw -1 and 1, the limit of a^n is zero.

    In particular, for a=2/3, we have that the limit of (2/3)^n is zero. Which is what you were wondering about.
     
  8. Nov 12, 2006 #7
    Ooh, ok, I understand now.

    Thanks!
     
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