1. Not finding help here? Sign up for a free 30min 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!

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?
     
  2. jcsd
  3. Nov 11, 2006 #2

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Limit of Sequence
  1. Sequences and Limits (Replies: 8)

  2. Limit of a Sequence (Replies: 18)

  3. Limits and Sequences (Replies: 8)

  4. Limit of a Sequence (Replies: 1)

Loading...