Prove Convergence of Sequence Defined by f(an) in R

In summary: It has the recursive rule ##r_{n+1} = f(r_n + L) - L## and the condition ##|r_{n+1}/r_n| < c##. So it looks like the original sequence ##a_n## except it is shifted by ##L##. Also, the condition ##0 < c < 1## is the condition for a contracting series. If a series is contracting, then it converges. So the statement is true.
  • #1
analysis001
21
0

Homework Statement


Consider the sequence {an}[itex]\subset[/itex]R which is recursively defined by an+1=f(an). Prove that if there is some L[itex]\in[/itex]R and a 0≤c<1 such that |[itex]\frac{a_{n+1}-L}{a_{n}-L}[/itex]|<c for all n[itex]\in[/itex]N then limn[itex]\rightarrow[/itex]∞an=L.


Homework Equations


Definition of convergence: Suppose (X,d) is a metric space and let {a_{n}}[itex]\subseteq[/itex]R denote a sequence in R. We will say {a[itex]_{n}[/itex]} converges in R if there is an L[itex]\in[/itex]R such that for every ε>0 there is an N'[itex]\in[/itex]N so that n'≥N' implies that d(a[itex]_{n}[/itex],L)<ε [itex]\Rightarrow[/itex] limn[itex]\rightarrow[/itex]∞an=L.


The Attempt at a Solution


Take ε>0. Since |an+1-L|< |an-L| [itex]\Rightarrow[/itex] |[itex]\frac{a_{n+1}-L}{a_{n}-L}[/itex]|<c. If n'=n+1 then N'=n. If 0<ε<|an+1-L| then d(an,L)=|an-L|<ε which by definition implies limn[itex]\rightarrow[/itex]∞an=L.

I'm not really sure if this is right. If anyone could tell me if anything is wrong with it that would be great! Thanks.
 
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  • #2
analysis001 said:

Homework Statement


Consider the sequence {an}[itex]\subset[/itex]R which is recursively defined by an+1=f(an). Prove that if there is some L[itex]\in[/itex]R and a 0≤c<1 such that |[itex]\frac{a_{n+1}-L}{a_{n}-L}[/itex]|<c for all n[itex]\in[/itex]N then limn[itex]\rightarrow[/itex]∞an=L.


Homework Equations


Definition of convergence: Suppose (X,d) is a metric space and let {a_{n}}[itex]\subseteq[/itex]R denote a sequence in R. We will say {a[itex]_{n}[/itex]} converges in R if there is an L[itex]\in[/itex]R such that for every ε>0 there is an N'[itex]\in[/itex]N so that n'≥N' implies that d(a[itex]_{n}[/itex],L)<ε [itex]\Rightarrow[/itex] limn[itex]\rightarrow[/itex]∞an=L.

You can give the definition of convergence of a sequence in a general metric space (X,d) and then immediately tell us that [itex]X = \mathbb{R}[/itex] and [itex]d: (x,y) \mapsto |x - y|[/itex], or you could just give the definition of convergence of a real sequence:

A real sequence [itex](a_n)[/itex] converges to [itex]L \in \mathbb{R}[/itex] if and only if for all [itex]\epsilon > 0[/itex] there exists an [itex]N \in \mathbb{N}[/itex] such that for all [itex]n \in \mathbb{N}[/itex] if [itex]n \geq N[/itex] then [itex]|a_n -L| < \epsilon[/itex] (or, equivalently, [itex]L - \epsilon < a_n < L + \epsilon[/itex]).

The Attempt at a Solution


Take ε>0. Since |an+1-L|< |an-L| [itex]\Rightarrow[/itex] |[itex]\frac{a_{n+1}-L}{a_{n}-L}[/itex]|<c.

You are given [itex]\left|\dfrac{a_{n+1}-L}{a_{n}-L}\right| < c < 1[/itex] for all [itex]n \in \mathbb{N}[/itex]. You conclude that [itex]|a_{n+1} - L| < |a_n - L|[/itex] for all [itex]n \in \mathbb{N}[/itex].

This tells you that [itex]|a_n - L|[/itex] is a strictly decreasing sequence bounded below by zero, and that it therefore converges to some [itex]S \geq 0[/itex].

You now need to show that [itex]S > 0[/itex] is impossible.
 
  • #3
analysis001 said:

Homework Statement


Consider the sequence {an}[itex]\subset[/itex]R which is recursively defined by an+1=f(an). Prove that if there is some L[itex]\in[/itex]R and a 0≤c<1 such that |[itex]\frac{a_{n+1}-L}{a_{n}-L}[/itex]|<c for all n[itex]\in[/itex]N then limn[itex]\rightarrow[/itex]∞an=L.


Homework Equations


Definition of convergence: Suppose (X,d) is a metric space and let {a_{n}}[itex]\subseteq[/itex]R denote a sequence in R. We will say {a[itex]_{n}[/itex]} converges in R if there is an L[itex]\in[/itex]R such that for every ε>0 there is an N'[itex]\in[/itex]N so that n'≥N' implies that d(a[itex]_{n}[/itex],L)<ε [itex]\Rightarrow[/itex] limn[itex]\rightarrow[/itex]∞an=L.


The Attempt at a Solution


Take ε>0. Since |an+1-L|< |an-L| [itex]\Rightarrow[/itex] |[itex]\frac{a_{n+1}-L}{a_{n}-L}[/itex]|<c. If n'=n+1 then N'=n. If 0<ε<|an+1-L| then d(an,L)=|an-L|<ε which by definition implies limn[itex]\rightarrow[/itex]∞an=L.

I'm not really sure if this is right. If anyone could tell me if anything is wrong with it that would be great! Thanks.

Look at the new sequence ##r_n = a_n - L##.
 

1. What does it mean for a sequence to converge?

Convergence of a sequence means that the terms in the sequence approach a specific value or limit as the number of terms in the sequence increases.

2. How do you prove convergence of a sequence?

To prove convergence of a sequence defined by f(an) in R, one must show that the limit of the sequence, as n approaches infinity, exists and is equal to a specific value. This can be done using mathematical techniques such as the squeeze theorem, the limit comparison test, or the ratio test.

3. What is the squeeze theorem?

The squeeze theorem, also known as the sandwich theorem, states that if two functions, g(x) and h(x), both approach the same limit as x approaches a certain value, and f(x) is always between g(x) and h(x), then the limit of f(x) as x approaches the same value is also equal to the shared limit of g(x) and h(x).

4. What is the limit comparison test?

The limit comparison test is a method for determining the convergence or divergence of a series by comparing it to a known convergent or divergent series. If the limit of the quotient between the two series is greater than 0 and less than infinity, then the two series have the same convergence properties.

5. What is the ratio test?

The ratio test is a method for determining the convergence or divergence of a series by taking the limit of the ratio between consecutive terms. If the limit is less than 1, the series converges, if it is greater than 1, the series diverges, and if it is equal to 1, the test is inconclusive and another method must be used.

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