# Cauchy sequence

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• Oct 23rd 2006, 01:03 PM
JaysFan31
Cauchy sequence
I know that (an) and (bn) are cauchy sequences. I need to prove or disprove that (an+bn) is also cauchy.

Thanks for any suggestions.
• Oct 23rd 2006, 04:48 PM
ThePerfectHacker
Quote:

Originally Posted by JaysFan31
I know that (an) and (bn) are cauchy sequences. I need to prove or disprove that (an+bn) is also cauchy.

Thanks for any suggestions.

The sequence, $\displaystyle \{ a_n \}$ is a Cauchy sequence. Thus, for $\displaystyle n,m\geq N_1$ we have,
$\displaystyle |a_n-a_m| < \epsilon$ for any $\displaystyle \epsilon>0$

The sequence, $\displaystyle \{ b_n\}$ is a Cauchy sequence. Thus, for $\displaystyle n,m \geq N_2$ we have, $\displaystyle |b_n-b_m|< \epsilon$ for any $\displaystyle \epsilon>0$

Consider the sequence, $\displaystyle \{a_n+b_n\}$. For any $\displaystyle \epsilon /2>0$ we can find $\displaystyle N_1,N_2$ satisfy those two conditions on top. Let $\displaystyle N=\max\{N_1,N_2\}$ then,
$\displaystyle |a_n-a_m|<\epsilon/2$ and,
$\displaystyle |b_n-b_m|<\epsilon/2$.
Addition yields,
$\displaystyle |a_n-a_m|+|b_n-b_m|<\epsilon$
Triangular inequality (Thanks to CaptainBlank for teaching this move to me).
We have,
$\displaystyle |a_n+b_n-(a_m+b_m)|<\epsilon$ for $\displaystyle n,m\geq N$.
Q.E.D.
• Oct 23rd 2006, 11:03 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
The sequence, $\displaystyle \{ a_n \}$ is a Cauchy sequence. Thus, for $\displaystyle n,m\geq N_1$ we have,
$\displaystyle |a_n-a_m| < \epsilon$ for any $\displaystyle \epsilon>0$

The sequence, $\displaystyle \{ b_n\}$ is a Cauchy sequence. Thus, for $\displaystyle n,m \geq N_2$ we have, $\displaystyle |b_n-b_m|< \epsilon$ for any $\displaystyle \epsilon>0$

Consider the sequence, $\displaystyle \{a_n+b_n\}$. For any $\displaystyle \epsilon /2>0$ we can find $\displaystyle N_1,N_2$ satisfy those two conditions on top. Let $\displaystyle N=\max\{N_1,N_2\}$ then,
$\displaystyle |a_n-a_m|<\epsilon/2$ and,
$\displaystyle |b_n-b_m|<\epsilon/2$.
Addition yields,
$\displaystyle |a_n-a_m|+|b_n-b_m|<\epsilon$
Triangular inequality (Thanks to CaptainBlank for teaching this move to me).
We have,
$\displaystyle |a_n+b_n-(a_m+b_m)|<\epsilon$ for $\displaystyle n,m\geq N$.
Q.E.D.

This needs to start something like:

As $\displaystyle \{a_n\}$ is a Cauchy sequence, for all $\displaystyle \epsilon>0$ there exists a $\displaystyle N_1$ such that for all $\displaystyle n,m>N_1$ $\displaystyle |a_n-a_m|<\epsilon$.

With something similar for $\displaystyle \{b_n\}, \epsilon$ and $\displaystyle N_2$.

Then the rest follows as ImPerfectHacker has shown.

RonL
• Oct 24th 2006, 05:59 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
This needs to start something like:

As $\displaystyle \{a_n\}$ is a Cauchy sequence, for all $\displaystyle \epsilon>0$ there exists a $\displaystyle N_1$ such that for all $\displaystyle n,m>N_1$ $\displaystyle |a_n-a_m|<\epsilon$.

With something similar for $\displaystyle \{b_n\}, \epsilon$ and $\displaystyle N_2$.

Then the rest follows as ImPerfectHacker has shown.

RonL

I know, I was too lazy to write out: "For all ... there exits ... such that... then..."
• Oct 24th 2006, 10:54 AM
JaysFan31
Using the same from above, is (an*bn) Cauchy?
• Oct 24th 2006, 01:03 PM
ThePerfectHacker
Quote:

Originally Posted by JaysFan31
Using the same from above, is (an*bn) Cauchy?

Actually I have a much simpler proof that I should have stated for both the problems.

A sequence is a Cauchy sequence if and only if it is convergent.

Using this fact we can easily show that an*bn and an+bn is a cauchy sequence.

If an and bn are Cauchy sequences then they are convergent sequences. Then an*bn and an+bn are convergent sequences. That means then an*bn and an+bn are Cauchy sequences.