Sequence Convergence Proof

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Homework Statement


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]



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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 $x_n$
 

Answers and Replies

  • #2
jbunniii
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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]
 
  • #3
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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!
 
  • #4
jbunniii
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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!

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]?
 
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  • #5
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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?
 
  • #6
jbunniii
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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?

Looks good to me. You can even call it "max" instead of "sup" since there are only N+1 numbers under consideration.
 

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