# Easy cauchy sequence prob

1. Oct 14, 2005

### 1800bigk

How would one prove that the sum of 2 cauchy sequences is cauchy? I said let e>0 and take 2 arbitrary cauchy sequences then
|Sn - St|<e/2 whenever n,t>N1 and |St - Sm|<e/2 whenever t,m >N2.

So

|Sn - Sm|=|Sn - St + St - Sm|<= |Sn - St|+|St - Sm|< e/2 + e/2 <= e

So n,m>max{N1, N2} imples |Sn - Sm|<e thus cauchy

Am I close or way off?

2. Oct 14, 2005

### 1800bigk

bump ttt bump

3. Oct 15, 2005

### Oxymoron

The method looks perfect to me.

However, a slight issue with your subscripts. When I did this problem I only needed two subscripts. Observe:

Take two arbitrary Cauchy sequences $\{x_n\}$ and $\{y_n\}$. Then for any $\epsilon > 0$ there exists $N_1 \in \mathbb{N}$ such that for any $n,m \geq N_1$ we have

$$|x_n - x_m| < \frac{\epsilon}{2}$$

and there also exists an $N_2 \in \mathbb{N}$ such that for any $n,m \geq N_2$ we have

$$|y_n - y_m| < \frac{\epsilon}{2}$$

Then, as you said, choose $N = \max\{N_1,N_2\}$. Then for any $n,m \geq N$ we have

$$|(x_n + y_n) - (x_m - y_m)| = |(x_n - x_m) + (y_n - y_m)| \leq |x_n-x_m| + |y_n - y_m| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

Therefore $\{x_n + y_n\}$ is a Cauchy sequence. $\square$

4. Oct 15, 2005

thank you!