# Thread: HELP with cauchy sequences

1. ## HELP with cauchy sequences

I am so confused with Cauchy Sequences. I was sick the day we learned about them. Here are two problems that I am absolutely unsure about. (just two of many)

1. Prove straight from the definition that $\displaystyle (n + \frac{(-1)^n}{n})$ is NOT a Cauchy Sequences

and

2. Let $\displaystyle (x_{n})$ be a Cauchy Sequence such that $\displaystyle x_{n}$ is an integer for every $\displaystyle n \in N$. Show that $\displaystyle (x_{n})$ is ultimately constant.

2. Originally Posted by mremwo
I am so confused with Cauchy Sequences. I was sick the day we learned about them. Here are two problems that I am absolutely unsure about. (just two of many)

1. Prove straight from the definition that $\displaystyle (n + \frac{(-1)^n}{n})$ is NOT a Cauchy Sequences

and

2. Let $\displaystyle (x_{n})$ be a Cauchy Sequence such that $\displaystyle x_{n}$ is an integer for every $\displaystyle n \in N$. Show that $\displaystyle (x_{n})$ is ultimately constant.

And that is why you don't miss maths classes in college/university unless your city has

been nuked or you're completely dead...honest!

1) We get that for or any $\displaystyle \displaystyle{n\in\mathbb{N}\,,\,\,\left|n+\frac{(-1)^n}{n}-\left((n+1)+\frac{(-1)^{n+1}}{n+1}\right)\right|=}$

$\displaystyle \displaystyle{\left\{\begin{array}{ll}\left|-1-\frac{1}{n}-\frac{1}{n+1}\right|=\frac{n^2+3n+1}{n(n+1)}&\mbox {, if n is odd}\\{}\\\left|-1+\frac{1}{n}+\frac{1}{n+1}\right|=\frac{n^2+n-1}{n(n+1)}}&\mbox{, if n is even}\end{array}\right.}$

It's easy to see that in both cases we get that the resulting fraction is greater than or

equal 0.5, so choosing $\displaystyle \epsilon=0.1$, say, and calling the sequence $\displaystyle a_n$ , we get that for any

n, $\displaystyle |a_n-a_{n+1}|\nless \epsilon$ , which contradicts the definition of Cauchy seq. (just take m = n+1)

2) For such a sequence, if we have different elements, then $\displaystyle |x_n-x_m|\geq 1$ , but as the

seq. is Cauchy take say $\displaystyle \displaystyle{\epsilon=\frac{1}{2}\Longrightarrow \exists N\in\mathbb{N}\,\,s.t.\,\,\forall\,n,m>N\,,\,\,|x_ n-x_m|<\epsilon\Longleftrightarrow |x_n-x_m|=0$...and end

now the argument.

Tonio

3. Originally Posted by tonio
And that is why you don't miss maths classes in college/university unless your city has

been nuked or you're completely dead...honest!
Remember the Mathematicians code: "Them's that fall behind are left behind" Arrrgh..

CB

4. Can you use the fact that "a sequence of real numbers is Cauchy if and only if it converges"?

If so, they you can show that (1) is NOT a Cauchty sequence by showing that it is not convergent. And you can do that by showing it is not bounded.

For (2) show that $\displaystyle |a_{n+1}- a_n|< \epsilon$, where $\displaystyle a_{n+1}$ and $\displaystyle a_n$ are integers, requires that $\displaystyle a_{n+1}= a_n$.