# Check my proof. Please :)

• Jun 25th 2013, 04:58 PM
UNLVRich
So I'm studying analysis on my own which has the disadvantage that there aren't many people capable of checking proofs in my neighborhood. So if you would be so kind, I would truly appreciate it! (Happy)

Problem:
Suppose that $\{a_n\}$ is a Cauchy sequence. Prove that $\{a_{n}^{2}\}$ is a Cauchy sequence.

I came up with two proofs.

Proof 1:
Since $\{a_n\}$ is a Cauchy sequence, and the real numbers are complete, $\lim_{n\rightarrow \infty}a_n=a$ for some $a\in\Re$. Since the limit of a product of convergent sequences is the product of the limits of the convergent sequences, we have

$\lim_{n\rightarrow \infty}$$a_{n}^{2}$$=$${\lim_{n\rightarrow \infty}a_{n}}$$^2=a^2$.

Since $\{a_{n}^{2}\}$ converges to a finite limit, it is a Cauchy sequence. QED.

Proof 2:
Since $\{a_n\}$ is a Cauchy sequence, it converges to a finite limit. Since it converges to a finite limit, it is bounded. That is, there exists a number $M$ so that $\mid a_n\mid \leq M$ for all $n$. Using the triangle inequality, we see that

$\mid a_n+a_m \mid \quad \leq \quad \mid a_n\mid+\mid a_m \mid \quad \leq \quad 2M$ for all $m$ and $n$.

Let $\varepsilon > 0$ be given. Since $\{a_n\}$ is a Cauchy sequence, there exists an $N$ so that $n\geq N$ and $m\geq N$ implies $\mid a_n-a_m \mid \leq \frac{\varepsilon}{2M}$. But this means that
$\mid a_n^2-a_m^2\mid \quad = \quad \mid a_n+a_m \mid \mid a_n-a_m\mid \quad \leq \quad 2M\mid a_n-a_m \mid \quad \leq \quad \varepsilon$.

Therefore, $\{a_n^2\}$ is a Cauchy sequence. QED.
• Jun 25th 2013, 07:52 PM
hollywood
Re: Check my proof. Please :)
Both of your proofs assume that you have a sequence of real numbers, but that isn't specified in the problem.

In proof 2, you only use the completeness of the real numbers to show that the sequence is bounded. I think you can show that without assuming completeness.

- Hollywood
• Jun 26th 2013, 08:30 AM
UNLVRich
Re: Check my proof. Please :)
Quote:

Both of your proofs assume that you have a sequence of real numbers, but that isn't specified in the problem.

In proof 2, you only use the completeness of the real numbers to show that the sequence is bounded. I think you can show that without assuming completeness.

- Hollywood
Hi Hollywood,

Thank you for taking time to look at my proofs. I really appreciate it.

I guess I should have prefaced the post with the fact that I'm studying real analysis, so it is implied that I'm dealing with a sequence of real numbers.

In proof 2, I used completeness to show the sequence converges. Which in turn I used to show the sequence is bounded.