Thread: Question concerning Cauchy sequences.

Hi everyone,
I have struggled with this question for far too long and I am hopeful someone can help.

Let {x_n} be a sequence in R. Let 0<r<1 and suppose |x_n+1-x_n|<rⁿ for all n in N. I need to show {x_n} converges.
I know this can be done by showing {x_n} is a Cauchy sequence. I think I am on the right track for the first few steps. However, I quickly get lost. Please help.

Originally Posted by larsh74

Let {x_n} be a sequence in R. Let 0<r<1 and suppose {x_n+1-x_n}<rⁿ for all n in N. I need to show {x_n} converges.
I know this can be done by showing {x_n} is a Cauchy sequence. I think I am on the right track for the first few steps

The way you have posted this means the statement is false.

Example: let $\displaystyle x_n=-n$. Now $\displaystyle \forall~n$ we have $\displaystyle -1=x_{n+1}-x_{n}<\left(\frac{1}{2}\right)^n$.
You know that $\displaystyle x_n=-n$ is not a Cauchy sequence.

Do you mean $\displaystyle \left| {{x_{n + 1}} - {x_n}} \right| < {r^n}~?$

Originally Posted by Plato

The way you have posted this means the statement is false.

Example: let $\displaystyle x_n=-n$. Now $\displaystyle \forall~n$ we have $\displaystyle -1=x_{n+1}-x_{n}<\left(\frac{1}{2}\right)^n$.
You know that $\displaystyle x_n=-n$ is not a Cauchy sequence.

Do you mean $\displaystyle \left| {{x_{n + 1}} - {x_n}} \right| < {r^n}~?$

Yes, I definitely typed that wrong. I changed the original post, but I am still struggling with this.

Hi,
Maybe this will help:
For m > n,
$\displaystyle |x_m-x_n|=|\sum_{k=n}^{m-1}x_{k+1}-x_k|\leq\sum_{k=n}^{m-1}|x_{k+1}-x_k|\leq\sum_{k=n}^{m-1}r^k$

Find a closed form for the last sum, and you should then be able to prove the sequence is a Cauchy sequence.

Isn't the sequence in the problem strictly smaller than the sequence of partial sums of a geometric series, since 0 < r < 1? Therefore it is bounded.

Edit: actually it isn't necessarily either strictly increasing or strictly decreasing, since all we know is the absolute value of the difference of terms. But the sequence still has to converge, see if you can use the aforementioned fact to finish the proof.

I think the most natural way is to go with SworD's line of thought, esp. if this question is for an advanced calculus/analysis course.

$\displaystyle \text{If we let }m=n+1, \text{ and using the triangle inequality know that, } \\ |x_m| - |x_{m-1}| \leq |x_m-x_{m-1}| <r^{m-1}\leqr^m \quad \forall m\in \mathbb{N} \implies |x_m|<r^m +|x_{m-1}|\implies \text{ boundedness and continue with proof... }$

A WLOG (without loss of generality) assumption that you could make is that the sequence is increasing hence monotonic, etc.

Hope this helps.

Update: You also need to mention the existence of a convergent subsequence due to the boundedness condition. Since you know that this sequence is Cauchy, what does a convergent subsequence mean for the whole sequence (knowing that the convergent subsequence is arbitrary)?

The hint we were given is this:

|x_m-x_n|=|x_m-x_m+1+x_m+1-x_m+2+...+x_n-1+x_n|
<=|x_m-x_m+1|+|x_m+1-x_m+2|+...+|x_n-1+x_n|
<=r^m+r^m+1+...+rⁿ
<=r^m(1+r+r²+...r^n-m)

It was given by another student and I'm not sure it is correct. If it is, I can finish up the proof. I'm just not convinced that this is the right direction to go with this.

What class is this for? I'm just curious about how much material (theorems, definitions, etc.) on sequences/convergence you're supposed to know.

If you can't invoke any theorems/lemmas, then use the hint and proceed.

Correction from my last post:
$\displaystyle |x_m| - |x_{m-1}| \leq |x_m-x_{m-1}|<r^{m-1} \leq r^m $

This is used to show boundedness. the hint follows and you'll prove convergence.

Hi,
The following attachment is the "standard" proof of your result:

Originally Posted by chen09

Correction from my last post:
$\displaystyle |x_m| - |x_{m-1}| \leq |x_m-x_{m-1}|<r^{m-1} \leq r^m $
This is used to show boundedness. the hint follows and you'll prove convergence.

This is a correction to the above. The standard proof is given in reply #10.
BUT nonetheless, note that if $\displaystyle 0<r<1$ then $\displaystyle 0<r^n<r^{n-1}$ if $\displaystyle n\in\mathbb{Z}^+$.

Thanks for all the replies. I have a proof that I think is correct, but it is not the same as the proof in reply #10. This is a Real Number Analysis course. We are free to use any theorems or definitions that we have covered in class or in the homework. But, our selection is somewhat limited at this point. Thanks again for all the helpful replies.