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

[Ozma]

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

Is this correct? I am not sure about the assumption [imath]m \ge n[/imath] 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 [imath](p_n)\to L[/imath] then [imath](\forall \alpha>0)(\exists N\in \mathbb{N}^+)[n\ge N \Rightarrow \left|p_n-L\right|<\alpha[/imath]. Definition of convergence
[imath]\left|p_n-p_m\right|\le\left|p_n-L\right|+\left|L-p_m\right|<\alpha+\alpha=2\alpha[/imath]


[imath][/imath][imath][/imath][imath][/imath]

 

[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: [imath]n~\&~m[/imath]. I also left it for you to supply those details.

 

[Ozma]

@pka: Let [imath]\epsilon>0[/imath]. Since [imath]p_n \to L[/imath], then there exists [imath]n_1 \in \mathbb{N}[/imath] such that [imath]n \ge n_1 \implies |p_n-L|<\frac{1}{2}\epsilon[/imath]; so, if [imath]n,m \ge n_1[/imath] it is [imath]|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[/imath]. 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 [imath]L \in \mathbb{R}[/imath]. Thanks

 

[Steven G]

@Steven G: You are right. I must specify that [imath]L \in \mathbb{R}[/imath]. 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 [imath]L[/imath], the problem only asks to show that the sequence is a Cauchy sequence (I specified [imath]L\in\mathbb{R}[/imath] because, otherwise, the proof would not work). In any case, it is [imath]L=1[/imath], because since [imath]n \to \infty[/imath] I can assume [imath]n \ne 0[/imath] and so [imath]\frac{n+1}{n-2}=\frac{1+\frac{1}{n}}{1-\frac{2}{n}}[/imath].

 

[pka]

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

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

[imath][/imath][imath][/imath][imath][/imath]

 

[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 [imath]\mathbb{R}[/imath] 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 [imath]\{p_n\}[/imath] is [imath]1[/imath], 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 [imath]L[/imath], secondly I don't see how saying explicitly what the limit of [imath]\{p_n\}[/imath] is can be useful in proving it is Cauchy (if I am constrained to use only the abstract definition of Cauchy sequence).