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

So, in summary, the sequence converges to √2 when working in the reals, but not when working in the rationals. It is also a Cauchy sequence in the reals, but not in the rationals.
  • #1
hb123
2
0

Homework Statement


Let {pn}n[itex]\in[/itex]P be a sequence such that pn is the decimal expansion of [itex]\sqrt{2}[/itex] 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?


Homework Equations


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


The Attempt at a Solution


I haven't gotten very far with this. Obviously, the sequence converges to p=[itex]\sqrt{2}[/itex]. Thus when working in the rationals, it doesn't converge and when working in the reals it does converge (since [itex]\sqrt{2}[/itex] 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.
 
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  • #2
hb123 said:

Homework Statement


Let {pn}n[itex]\in[/itex]P be a sequence such that pn is the decimal expansion of [itex]\sqrt{2}[/itex] 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?


Homework Equations


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


The Attempt at a Solution


I haven't gotten very far with this. Obviously, the sequence converges to p=[itex]\sqrt{2}[/itex]. Thus when working in the rationals, it doesn't converge and when working in the reals it does converge (since [itex]\sqrt{2}[/itex] 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.

Pick a specific ε, say ε=1/10000. How large does N have to be?
 
  • #3
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| [itex]\geq[/itex] ϵ). Just to clarify, P is the set of all positive integers.
 
  • #4
hb123 said:
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| [itex]\geq[/itex] ϵ). Just to clarify, P is the set of all positive integers.

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.
 

1. What is the definition of a convergent sequence?

A sequence is said to be convergent if it approaches a certain value as the number of terms increases, and this value is called the limit of the sequence. In other words, the elements in a convergent sequence get closer and closer to a specific value as the sequence progresses.

2. How do you determine if a sequence is convergent?

To determine if a sequence is convergent, we can use the limit comparison test. This involves finding the limit of the sequence as the number of terms approaches infinity. If the limit equals a finite number, then the sequence is convergent. If the limit is infinity or does not exist, then the sequence is divergent.

3. What is the difference between a convergent sequence and a Cauchy sequence?

A convergent sequence approaches a specific value as the number of terms increases, while a Cauchy sequence has elements that get closer and closer to each other as the sequence progresses. In other words, a Cauchy sequence is a special type of convergent sequence where the elements become arbitrarily close to each other, rather than approaching a specific value.

4. How can you determine if a sequence is a Cauchy sequence?

To determine if a sequence is a Cauchy sequence, we can use the Cauchy criterion. This states that a sequence is Cauchy if the distance between any two elements in the sequence becomes arbitrarily small as the sequence progresses. In mathematical terms, this means that for any positive number ε, there exists a positive integer N such that for all n, m ≥ N, the absolute value of the difference between the nth and mth terms is less than ε.

5. Can a sequence be both convergent and Cauchy?

Yes, a sequence can be both convergent and Cauchy. In fact, all convergent sequences are also Cauchy sequences. This is because as a sequence approaches a specific value, its elements also become arbitrarily close to each other. However, not all Cauchy sequences are convergent, as there may not be a specific limit that the elements are approaching.

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