# {a_n} converges => {a_n} is Cauchy; brief question

• October 20th 2009, 05:49 PM
cgiulz
{a_n} converges => {a_n} is Cauchy; brief question
Prove. {a_n} converges, then it is Cauchy.

I think I already got the answer. Can someone let me know if I'm allowed to start like this.

|a_n - L|< e for n>N, and
|a_m - L|< e for m>N

|a_n - a_m| = |(a_n - L) - (a_m - L)|...
• October 20th 2009, 05:52 PM
Defunkt
Quote:

Originally Posted by cgiulz
Prove. {a_n} converges, then it is Cauchy.

I think I already got the answer. Can someone let me know if I'm allowed to start like this.

|a_n - L|< e for n>N, and
|a_m - L|< e for m>N

|a_n - a_m| = |(a_n - L) - (a_m - L)|...

... $= |(a_n-L) + (L-a_m)| \leq |a_n-L| + |a_m - L| < \epsilon + \epsilon$... yes, it is indeed correct! (Clapping)