Show that (n+1)/(n-2) is Cauchy using the definition

[Ozma]

I have to show that the sequence $\{p_n\}$ defined by $p_n=\frac{n+1}{n-2}$ for $n=3,4,\dots$ is a Cauchy sequence using the definition. I tried this:
$$|p_n-p_m|=\left|\frac{n+1}{n-2}-\frac{m+1}{m -2}\right|=\left|\frac{-3n+3m}{(n-2)(m-2)}\right|=\frac{3|m-n|}{(n-2)(m+2)} ..............edited[COLOR=rgb(0, 0, 0)]$$[/COLOR]
Assuming $m\ge n$, it is
$$\frac{3|m-n|}{(n-2)(m+2)}=\frac{3(m-n)}{(n-2)(m-2)}=3\frac{m-2+2-n}{(n-2)(m-2)}=\frac{3}{n-2}-\frac{3}{m-2}$$Since $-\frac{3}{m-2}<0$ because $m=3,4,\dots$, it is
$$\frac{3}{n-2}-\frac{3}{m-2}<\frac{3}{n-2}$$Since $\frac{3}{n-2} \to 0$ as $n \to \infty$, for any $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $n,m \ge N$ implies $|p_n-p_m|<\epsilon$.

Is this correct? I am not sure about the assumption $m \ge n$ and the inequalities I used.

 

[Steven G]

Your 1st line is wrong........................fixed

 

[Steven G]

Note that |pm -pn| = |pn-pm|

 

[Steven G]

It looks good to me. I would have said that 3/(m-2)>0 instead of -3/(m-2)<0 but they are the same.
There is no problem saying that m≥n, since |pm -pn| = |pn-pm| anyways.

 

[pka]

If $(p_n)\to L$ then $(\forall \alpha>0)(\exists N\in \mathbb{N}^+)[n\ge N \Rightarrow \left|p_n-L\right|<\alpha$. Definition of convergence
$\left|p_n-p_m\right|\le\left|p_n-L\right|+\left|L-p_m\right|<\alpha+\alpha=2\alpha$


$$$$$$

 

[Ozma]

@Steven G Yes there was a typo, thanks for noticing it and thanks for the help! Thanks to the moderator who fixed it too.

@pka: Oh nice solution, thanks. I would have used a similar approach I believe (using the fact that a sequence that converges in a complete space is Cauchy) but I think that my lecturer wanted to make us practice with two indexes.

 

[pka]

I used two indexes: $n~\&~m$. I also left it for you to supply those details.

 

[Ozma]

@pka: Let $\epsilon>0$. Since $p_n \to L$, then there exists $n_1 \in \mathbb{N}$ such that $n \ge n_1 \implies |p_n-L|<\frac{1}{2}\epsilon$; so, if $n,m \ge n_1$ it is $|p_n-p_m|=|p_n-L+L-p_m|\le|p_n-L|+|p_m-L|<\frac{1}{2}\epsilon+\frac{1}{2}\epsilon=\epsilon$. You meant this details? Thanks for the suggestion.

About the two indexes, sorry, I meant: "working with two explicit sequences and make practice with quantities depending on two indexes".

 

[Steven G]

I would not accept that proof as being complete as you are being secretive as to what the value of L is.

 

[Ozma]

@Steven G: You are right. I must specify that $L \in \mathbb{R}$. Thanks

 

[Steven G]

@Steven G: You are right. I must specify that $L \in \mathbb{R}$. Thanks

No, you should state the exact value of L and stop being secretive about its value. Yes, L is a real number and it equals ....

 

[Ozma]

@Steven G: I don't completely get the reason why we should evaluate $L$, the problem only asks to show that the sequence is a Cauchy sequence (I specified $L\in\mathbb{R}$ because, otherwise, the proof would not work). In any case, it is $L=1$, because since $n \to \infty$ I can assume $n \ne 0$ and so $\frac{n+1}{n-2}=\frac{1+\frac{1}{n}}{1-\frac{2}{n}}$.

 

[pka]

@Steven G: I don't completely get the reason why we should evaluate $\bf\large L$,

I agree. the sequence $\left(p_n\right)$ is a sequence of real that converges to $L$ so what else would $L$ be if not real?

$$$$$$

 

[Steven G]

If you are going to bother to say that the sequences converges to L, then why not just say that the sequences converges to 1??

 

[Ozma]

@Steven G: I was following the hint from pka, so I was coherent with its notation; I don't understand your point. I could even say that, since $\mathbb{R}$ is complete, every sequence of real numbers is convergent if and only if it is Cauchy and so, since it is easy to show that the limit of $\{p_n\}$ is $1$, solve the problem in this way. But this wasn't the aim of this exercise, as I said the definition of Cauchy sequence must be used and: firstly, the definition is given with a generic limit $L$, secondly I don't see how saying explicitly what the limit of $\{p_n\}$ is can be useful in proving it is Cauchy (if I am constrained to use only the abstract definition of Cauchy sequence).