Proving the Cauchy Criterion for Sum of Sequences

[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?

 

[1800bigk]

bump ttt bump

 

[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 &gt; 0 there exists N_1 \in \mathbb{N} such that for any n,m \geq N_1 we have

<br /> |x_n - x_m| &lt; \frac{\epsilon}{2}<br />

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

<br /> |y_n - y_m| &lt; \frac{\epsilon}{2}<br />

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| &lt; \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon

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

 

[1800bigk]

thank you!