# Determining if a sequence is convergent and/or a Cauchy sequence

1. Nov 12, 2012

### hb123

1. The problem statement, all variables and given/known data
Let {pn}n$\in$P be a sequence such that pn is the decimal expansion of $\sqrt{2}$ truncated after the nth decimal place.
a) When we're working in the rationals is the sequence convergent and is it a Cauchy sequence?
b) When we're working in the reals is the sequence convergent and is it a Cauchy sequence?

2. Relevant equations
A sequence {pn}n$\in$P converges to p if ($\forall\epsilon>0$)($\exists N \in P$)($\forall n\geq N$)(| pn-p| < $\epsilon$).
It is a Cauchy sequence if ($\forall\epsilon>0$)($\exists N \in P$)($\forall n,m\geq N$)(|pn-pm|< $\epsilon$).

3. The attempt at a solution
I haven't gotten very far with this. Obviously, the sequence converges to p=$\sqrt{2}$. Thus when working in the rationals, it doesn't converge and when working in the reals it does converge (since $\sqrt{2}$ is a real number but not rational). However, I'm stuck trying to prove this using the mentioned definitions, and can't get anywhere with trying to prove if the sequence is Cauchy.

2. Nov 12, 2012

### Dick

Pick a specific ε, say ε=1/10000. How large does N have to be?

3. Nov 12, 2012

### hb123

The thing is, though, I need to show that this is true for all ε, not just an arbitrary one. On the other hand, if I'm trying to prove that the series doesn't converge to a p, then I'd need to show that !(∀ϵ>0)(∃N∈P)(∀n≥N)(| pn-p| < ϵ), or (∃ϵ>0)(∀N∈P)(∃≥N)(|pn-p| $\geq$ ϵ). Just to clarify, P is the set of all positive integers.

4. Nov 12, 2012

### Dick

I know. I'm suggesting you think about the N corresponding to specific case ε=0.0001 to get you started. Working it out is really not much different from a general value of ε. And I'm talking about the Cauchy part of the proof here.