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

Cauchy Sequence

  • Thread starter tylerc1991
  • Start date
  • #1
166
0

Homework Statement



Show that the sequence of real numbers defined by [itex]x_{n + 1} = x_n + \frac{1}{x_n^2}, \, x_1 = 1[/itex] is not a Cauchy sequence.

Homework Equations



A sequence [itex]\{ p_n \}[/itex] is Cauchy if and only if, for all [itex]\varepsilon > 0[/itex], there exists an [itex]N > 0[/itex] such that [itex]d(p_n, p_m) < \varepsilon[/itex] for all [itex]m, n > N[/itex].

The Attempt at a Solution



We can assume that [itex]d[/itex] is the usual metric on [itex]\mathbb{R}[/itex]. I don't even see where to begin. I see that the sequence is monotonically increasing, so that
[itex]1 = \frac{1}{x_1} > \frac{1}{x_2} > \frac{1}{x_3} > \dotsb.[/itex]
So
[itex]1 = \frac{1}{x_1^2} > \frac{1}{x_2^2} > \frac{1}{x_3^2} > \dotsb.[/itex]
To me it looks like the sequence is in fact Cauchy. Please help!
 

Answers and Replies

  • #2
161
0
Well if you're trying to show that it is not Cauchy, state what it means for a sequence to not be Cauchy. That is where I would start :)
 
  • #3
166
0
Well if you're trying to show that it is not Cauchy, state what it means for a sequence to not be Cauchy. That is where I would start :)
The sequence isn't Cauchy if there exists an [itex]\varepsilon > 0[/itex] such that for all [itex]N > 0[/itex] there exists [itex]m, n > 0[/itex] such that [itex]d(x_m, x_n) \geq \varepsilon[/itex].

...

still stuck

...
 
  • #4
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,508
730
Do you have the theorem that a sequence is Cauchy if and only if it is convergent? And if so, what happens if you suppose ##\lim_{x\rightarrow \infty} = L##?
 
  • #5
166
0
Do you have the theorem that a sequence is Cauchy if and only if it is convergent? And if so, what happens if you suppose ##\lim_{x\rightarrow \infty} = L##?
Yes, I can assume the sequence is Cauchy if and only if it is convergent. The definition of limits that we are using states that [itex]\lim_{n \to \infty} x_n = L[/itex] if and only if
[itex]\forall \varepsilon > 0 \, \exists N > 0 \, s.t. \, \forall n > N \quad d(L, x_n) < \varepsilon[/itex].

Now can I somehow use the fact that [itex]x_n[/itex] is increasing to say that [itex]d(L, x_n)[/itex] is always increasing? And hence it is greater than or equal to [itex]\varepsilon[/itex] for some [itex]n[/itex]?
 
  • #6
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,508
730
Yes, I can assume the sequence is Cauchy if and only if it is convergent. The definition of limits that we are using states that [itex]\lim_{n \to \infty} x_n = L[/itex] if and only if
[itex]\forall \varepsilon > 0 \, \exists N > 0 \, s.t. \, \forall n > N \quad d(L, x_n) < \varepsilon[/itex].

Now can I somehow use the fact that [itex]x_n[/itex] is increasing to say that [itex]d(L, x_n)[/itex] is always increasing? And hence it is greater than or equal to [itex]\varepsilon[/itex] for some [itex]n[/itex]?
It's easier than that. If the sequence has a limit, what happens if you take the limit of both sides of your recursion?
 
  • #7
166
0
It's easier than that. If the sequence has a limit, what happens if you take the limit of both sides of your recursion?
The limit of both sides of the recurrence should then equal the same thing, namely [itex]L[/itex]. Then I would have that

[itex]\lim_{n \to \infty} x_{n + 1} = \lim_{n \to \infty} x_n + \frac{1}{x_n^2} = L + \lim_{n \to \infty} \frac{1}{x_n^2} = L[/itex].

Doesn't this simply prove that [itex]\lim_{n \to \infty} \frac{1}{x_n^2} = 0[/itex]?
 
  • #8
166
0
The limit of both sides of the recurrence should then equal the same thing, namely [itex]L[/itex]. Then I would have that

[itex]\lim_{n \to \infty} x_{n + 1} = \lim_{n \to \infty} x_n + \frac{1}{x_n^2} = L + \lim_{n \to \infty} \frac{1}{x_n^2} = L[/itex].

Doesn't this simply prove that [itex]\lim_{n \to \infty} \frac{1}{x_n^2} = 0[/itex]?
But then doesn't this imply [itex]\lim_{n \to \infty} x_n^2 = \infty[/itex]? completing the problem?
 
  • #9
Matterwave
Science Advisor
Gold Member
3,965
326
Can that second limit be 0 while the first limit exists? Is there a number L such that 1/L^2=0?
 
  • #10
166
0
Can that second limit be 0 while the first limit exists? Is there a number L such that 1/L^2=0?
Exactly. Thank you LCKurtz for your patience! :)
 

Related Threads for: Cauchy Sequence

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
1
Views
858
  • Last Post
Replies
2
Views
657
  • Last Post
Replies
17
Views
3K
Replies
1
Views
1K
Top