1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Sequence Convergence Proof

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Let [tex]X=(x_n) [/tex] be a sequence of strictly positive numbers such that [tex]\lim(x_{n+1}/x_n)<1[/tex]. Show for some [tex]0<r<1[/tex], and for some [tex]C>0[/tex], [tex]0<x_n<Cr^n[/tex]



    2. Relevant equations



    3. The attempt at a solution
    Let [tex]\lim(x_{n+1}/x_n)=x<1[/tex]
    By definition of the limit, [tex]\lim(x_{n+1}/x_n)=x \Rightarrow \forall \epsilon>0 [/tex] there exists [tex] \: K(\epsilon) [/tex] such that [tex]. \: \forall n>K(\epsilon) [/tex]

    [tex]|\frac{x_{n+1}}{x_n}-x|<\epsilon[/tex]
    Since i can pick any epsilon, let epsilon be such that [tex] \epsilon + x = r <1[/tex]. Also, I know that since this is a positive sequence, [tex]\frac{x_{n+1}}{x_n}>0[/tex]. Therefore, for large enough [tex]n[/tex],

    [tex]0<\frac{x_{n+1}}{x_n}<r<1. [/tex]

    From here I am not sure where to go, any hints would be much appreciated! I cannot find out what this tells me about $x_n$
     
  2. jcsd
  3. Mar 22, 2010 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    For large enough [itex]n[/itex] you also have the following:

    [tex]\frac{x_{n+k}}{x_n} = \frac{x_{n+1}}{x_n} \frac{x_{n+2}}{x_{n+1}} \cdots \frac{x_{n+k}}{x_{n+k-1}} < r^k[/tex]
     
  4. Mar 22, 2010 #3
    Can I let [tex]x_n=C[/tex] and therefore say the sequence [tex] x_{n+k}<Cr^k[/tex]? If so then n would fixed and k would be the index, correct? Thanks for the speedy response!
     
  5. Mar 22, 2010 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You're not quite there yet.

    First note that you need

    [tex]x_{n} < C r^n[/tex]

    not

    [tex]x_{n+k} < Cr^k[/tex]

    Also, it needs to be true for EVERY [itex]n[/itex], not just sufficiently large [itex]n[/itex].

    First let's introduce some notation. It is easier to discuss if we give this "sufficiently large" [itex]n[/itex] a name, say [itex]N[/itex]. For all [itex]n \geq N[/itex],

    [tex]0 < \frac{x_{n+1}}{x_n} < r < 1[/tex]

    which yields

    [tex]x_{N+k} < r^k x_N[/tex]

    for all [tex]k \geq 0[/tex].

    I can get the index and exponent to agree by writing, equivalently,

    [tex]x_{N+k} < r^{N+k} (x_N r^{-N})[/tex]

    Then if I set [tex]C_1 = x_N r^{-N}[/tex] I get

    [tex]x_{N+k} < C_1 r^{N+k}[/tex]

    which looks pretty promising. However, this [itex]C_1[/itex] may not be large enough to work for ALL [itex]n[/itex], i.e. it may not be true that

    [tex]x_n < C_1 r^n[/tex]

    for all [itex]n < N[/itex].

    But note that there are only finitely many [itex]x_n[/itex] with [itex]n < N[/itex]. Can you use that fact to find a [itex]C[/itex] that does work for all [itex]n[/itex]?
     
    Last edited: Mar 22, 2010
  6. Mar 22, 2010 #5
    So, for every [tex]j<N[/tex], Let [tex]M_j[/tex] be such that [tex]x_j<M_jr^j[/tex] (there exists such an M by the archimedean property). Now, if I take [tex]C=\sup\{M_j,C_1| j\in\mathbb{N},j<N\}[/tex], that should do the job since there are only finitely many M's, correct?
     
  7. Mar 22, 2010 #6

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Looks good to me. You can even call it "max" instead of "sup" since there are only N+1 numbers under consideration.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook