• Support PF! Buy your school textbooks, materials and every day products Here!

Cauchy sequence

  • Thread starter Oxymoron
  • Start date
  • #1
870
0
Question

Consider the sequence [itex]\{p^n\}_{n\in\mathbb{N}}[/itex]. Prove that this sequence is Cauchy with respect to the p-adic metric on [itex]\mathbb{Q}[/itex]. What is the limit of the sequence?
 

Answers and Replies

  • #2
870
0
Solution

Let [itex]p_n = 1 + p + p^2 + \dots + p^{n-1}[/itex]. Then we have

[tex]|p_{n+k}-p_n|_p = \left|p^n + p^{n+1} + \dots + p^{n+k-1}\right|_p[/tex]
[tex]= \left|p^n(1+p+p^2 + \dots + p^{k-1})\right|_p[/tex]
[tex]= \frac{1}{p^n}[/tex]

So for any [itex]\epsilon > 0[/itex], we can choose an [itex]N\in\mathbb{N}[/itex] such that [itex]p^N \geq \frac{1}{\epsilon}[/itex], so if [itex]n > N[/itex] we have

[tex]|p_{n+k} - p_n|_p < \frac{1}{p^N} \leq \epsilon[/tex]

Therefore [itex]\{p^n\}_{n\in\mathbb{N}}[/itex] is Cauchy.
 
  • #3
870
0
ii) Since

[tex]|p^n|_p = \frac{1}{p^n} \rightarrow 0 \quad \mbox{as} \quad n\rightarrow \infty[/tex]

the limit

[tex]\lim_{n\rightarrow\infty}^p p^n = 0[/tex]

Hence this sequence is actually a null sequence with respect to the p-adic norm.
 
  • #4
shmoe
Science Advisor
Homework Helper
1,992
1
You've shown the corresponding sequence of partial sums is Cauchy, not the sequence itself. In your other p-adic thread you showed the terms, not the partial sums, go to zero. You seem to be mixing the sequence itself with the sequence of partial sums (or you swapped your replies).
 

Related Threads on Cauchy sequence

  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
2
Views
1K
Replies
21
Views
5K
Replies
1
Views
5K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
6
Views
2K
Replies
16
Views
12K
Replies
4
Views
31K
  • Last Post
Replies
1
Views
1K
Top