# Check my proof. Please :)

• June 25th 2013, 05:58 PM
UNLVRich
Check my proof. Please :)
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.
• June 25th 2013, 08:52 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
• June 26th 2013, 09:30 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.