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

[analysis001]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

Take ε>0. Since |a_n+1-L|< |a_n-L| $\Rightarrow$ |$\frac{a_{n+1}-L}{a_{n}-L}$|<c. If n'=n+1 then N'=n. If 0<ε<|a_n+1-L| then d(a_n,L)=|a_n-L|<ε which by definition implies lim_{n$\rightarrow$∞}a_n=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.

 

[pasmith]

Homework Statement

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

Homework Equations

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

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

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

The Attempt at a Solution

Take ε>0. Since |a_n+1-L|< |a_n-L| $\Rightarrow$ |$\frac{a_{n+1}-L}{a_{n}-L}$|<c.

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

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

You now need to show that $S > 0$ is impossible.

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

Take ε>0. Since |a_n+1-L|< |a_n-L| $\Rightarrow$ |$\frac{a_{n+1}-L}{a_{n}-L}$|<c. If n'=n+1 then N'=n. If 0<ε<|a_n+1-L| then d(a_n,L)=|a_n-L|<ε which by definition implies lim_{n$\rightarrow$∞}a_n=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 $r_n = a_n - L$.