1. ## 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.

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

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

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

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

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

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

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

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

Then the rest follows as ImPerfectHacker has shown.

RonL

4. Originally Posted by CaptainBlack
This needs to start something like:

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

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

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..."

5. Using the same from above, is (an*bn) Cauchy?

6. 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.