# Thread: Cauchy sequence and its limit......

1. ## Cauchy sequence and its limit......

Hi, I had difficulty solving this problem. Actually, I'm stuck.

Suppose (Xn) in R is defined recursively by X1= 1 and $\displaystyle x_{n+1}=\dfrac{1}{3+x_n}$, n= 1, 2, ...
Show (Xn) converges and find its limit.

And there's a hint: Notice that | Xn+2 - X n+1| < 1/9 |Xn - Xn+1|. Show (Xn) is Cauchy.

Thanks for any help.

I edited my mistake. Sorry again.

AS posted, that sequence is not bounded.
So it must be wrong.

3. Thank you, but why is it not bounded? I could not understand.

4. Originally Posted by truevein
Thank you, but why is it not bounded? I could not understand.
If we add $\frac{1}{3}$ enough times we get a large number.
The way you have it every third term of the sequence is a positive integer.

5. Sorry, it is not (1/3)+ Xn but 1/(3+Xn). It is bounded, isn't it?

6. Hint:

Induction (to the Q. hint)

7. Originally Posted by truevein
Sorry, it is not (1/3)+ Xn but 1/(3+Xn). It is bounded, isn't it?
Why not learn to post in symbols? You can use LaTeX tags
$$\displaystyle x_{n+1}=\dfrac{1}{3+x_n}$$ gives $\displaystyle x_{n+1}=\dfrac{1}{3+x_n}$
At least learn to use grouping symbols.
That wasted time and effort.

8. Originally Posted by truevein
Hi, I had difficulty solving this problem. Actually, I'm stuck.

Suppose (Xn) in R is defined recursively by X1= 1 and $\displaystyle x_{n+1}=\dfrac{1}{3+x_n}$, n= 1, 2, ...
Show (Xn) converges and find its limit.

And there's a hint: Notice that | Xn+2 - X n+1| < 1/9 |Xn - Xn+1|. Show (Xn) is Cauchy.

Thanks for any help.

I edited my mistake. Sorry again.
In terms of difference equation the sequence is the solution of the recursive equation...

$\displaystyle \Delta_{n} = x_{n+1}-x_{n} = \frac{1-3\ x_{n} - x_{n}^{2}}{3+x_{n}} = f(x_{n})$ (1)

The function $f(*)$ is represented here...

It exists one attractive fixed point in $x_{0} = \frac{-3+\sqrt{13}}{2}= .3027756377199...$ and for any $x_{1} > -3$ the sequence will converge to $x_{0}$. An interesting detail: because is $|f(x)|> |x_{0} - x|$ the sequence will be 'oscillating'. That is confirmed for $x_{1}=1$...

$x_{1}=1, x_{2}= .25, x_{3}= .307692..., x_{4}= .302325..., x_{5}= .3028169..., ...$

For $x_{1}< -3$ the behavior of the sequence is stiil to be analysed...

A very interesting question! ...

Kind regards

$\chi$ $\sigma$

9. In this post I will try to prove something more general.

Let $a_n$ be a sequence. Suppose there exist $1>\theta \in \mathbb{R}$ so that $\left | a_{n+1}-a_n \right |<\theta \left | a_n-a_{n-1} \right |$ for all $n\geq 2$. I will prove that $a_n$ converges.

Proof:

We prove via induction that for all $n\in \mathbb{N}$: $\left | a_{n+1}-a_n \right |\leq \left | \theta \right | ^{n-1}\left | a_2-a_1 \right |$.

Base of induction:

Of course: $\left | a_2-a_1 \right |\leq \left | \theta \right |^0\left | a_2-a_1 \right |$

Step of induction:

Let be $k\in \mathbb{N}$such that: $\left | a_{k+1}-a_k \right |\leq \left | \theta \right | ^{k-1}\left | a_2-a_1 \right |$, hence:

$\left | a_{k+2}-a_{k+1} \right |<\left | \theta \right |\left | a_{k+1}-a_k \right |\leq \left | \theta \left |\left | \theta \right |^{k-1}\left | a_2-a_1 \right |=\left | \theta \right |^k\left | a_2-a_1 \right |$.

...hence the statement is true for all $n\in \mathbb{N}$ .

Now, for all $m,n \in \mathbb{N}$ that $n>m$ we have:

$\displaystyle\left | a_n-a_m \right |=\left | \sum_{k=m}^{n-1}(a_{k+1}-a_k) \right |\leq \sum_{k=m}^{n-1}\left |a_{k+1}-a_k \right |< \left | a_2-a_1 \right |\sum_{k=m}^{n-1}\left | \theta \right |^{k-1}=$

$\displaystyle= \left | a_2-a_1 \right |\left | \theta \right |^{m-1}\sum_{r=0}^{n-1-m}\left | \theta \right |^r\leq \left | a_2-a_1 \right | \left | \theta \right |^{m-1}\frac{1-\left |\theta \right | ^{n-m}}{1-\left ( \theta \right )}\leq \left | a_2-a_1 \right |\left | \theta \right |^{m-1}\frac{1}{1-\left | \theta \right |}\overset{m\rightarrow \infty}{\rightarrow}0$

Hence, $a_n$ is Cauchy sequence, and it converges.

10. Thanks, chisigma and Also sprach Zarathustra, for your time and effort.
It looks a bit complicated, I'm trying to figure out and I will. Thanks again!