Proving the Sequence of Real Numbers is Not Cauchy

[tylerc1991]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

 

[Kindayr]

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 :)

 

[tylerc1991]

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 \varepsilon > 0 such that for all N > 0 there exists m, n > 0 such that d(x_m, x_n) \geq \varepsilon.

...

still stuck

...

 

[LCKurtz]

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 $\lim_{x\rightarrow \infty} = L$?

 

[tylerc1991]

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 \lim_{n \to \infty} x_n = L if and only if
\forall \varepsilon > 0 \, \exists N > 0 \, s.t. \, \forall n > N \quad d(L, x_n) < \varepsilon.

Now can I somehow use the fact that x_n is increasing to say that d(L, x_n) is always increasing? And hence it is greater than or equal to \varepsilon for some n?

 

[LCKurtz]

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 \lim_{n \to \infty} x_n = L if and only if
\forall \varepsilon > 0 \, \exists N > 0 \, s.t. \, \forall n > N \quad d(L, x_n) < \varepsilon.

Now can I somehow use the fact that x_n is increasing to say that d(L, x_n) is always increasing? And hence it is greater than or equal to \varepsilon for some n?

It's easier than that. If the sequence has a limit, what happens if you take the limit of both sides of your recursion?

 

[tylerc1991]

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 L. Then I would have that

\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.

Doesn't this simply prove that \lim_{n \to \infty} \frac{1}{x_n^2} = 0?

 

[tylerc1991]

The limit of both sides of the recurrence should then equal the same thing, namely L. Then I would have that

\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.

Doesn't this simply prove that \lim_{n \to \infty} \frac{1}{x_n^2} = 0?

But then doesn't this imply \lim_{n \to \infty} x_n^2 = \infty? completing the problem?

 

[Matterwave]

Can that second limit be 0 while the first limit exists? Is there a number L such that 1/L^2=0?

 

[tylerc1991]

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! :)