Sequence Convergence Proof

In summary, for a sequence X=(x_n) of strictly positive numbers with a limit of \lim(x_{n+1}/x_n)<1, there exists some 0<r<1 and some C>0 such that 0<x_n<Cr^n for all n in the sequence. This can be shown by setting C=max{M_j,C_1| j\in\mathbb{N},j<N} where M_j is such that x_j<M_jr^j and C_1=x_Nr^{-N}.
  • #1
blinktx411
35
0

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]



Homework Equations





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 $x_n$
 
Physics news on Phys.org
  • #2
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
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
blinktx411 said:
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]?
 
Last edited:
  • #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?
 
  • #6
blinktx411 said:
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.
 

1. What is a sequence convergence proof?

A sequence convergence proof is a mathematical method used to show that a sequence of numbers approaches a specific limit as the number of terms in the sequence increases. It is commonly used in calculus and other areas of mathematics to prove the convergence of infinite series.

2. How is a sequence convergence proof different from a limit proof?

A sequence convergence proof focuses on showing that the terms in a sequence get closer and closer to a specific limit as the number of terms increases, while a limit proof focuses on showing that a function approaches a specific value as the input approaches a certain value. However, both methods involve showing that a value is approached or reached as the number of terms or the input increases.

3. What are the main steps involved in a sequence convergence proof?

The main steps in a sequence convergence proof are choosing a limit to prove, showing that the terms in the sequence get closer and closer to that limit, and using mathematical tools such as the epsilon-delta definition of a limit or the squeeze theorem to formally prove the convergence of the sequence.

4. What are some common techniques used in sequence convergence proofs?

Some common techniques used in sequence convergence proofs include the comparison test, the ratio test, the root test, and the integral test. These methods involve comparing the given sequence to a known convergent or divergent sequence, using algebraic manipulation, or using calculus concepts to prove convergence.

5. Why is it important to prove the convergence of a sequence?

Proving the convergence of a sequence is important in mathematics because it allows us to make accurate predictions about the behavior of the sequence in the long run. It also helps us understand the behavior of functions and infinite series, which have a wide range of applications in various fields such as physics, engineering, and economics.

Similar threads

Replies
1
Views
531
  • Calculus and Beyond Homework Help
Replies
13
Views
899
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
14
Views
428
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
792
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
934
Back
Top