# Can you prove the following Cauchy sequence

Let (An) be a Cauchy sequence such that n is a member of N, and let c be a member of R, prove that (c*An) is also a Cauchy sequence

• In the definition take $N$ s.t. $|A_n-A_m|<\frac{\varepsilon}{|c|}$ when $n,m>N$ and the claim follow. – user386627 Nov 3 '17 at 12:27
• And if $c<0$ ??? – Fred Nov 3 '17 at 12:29
• @Fred: I just gave a comment, not an answer... The OP can adapt properly I'm sure ;-) By the way, in what your answer give something more than my comment ? – user386627 Nov 3 '17 at 12:30

## 1 Answer

If $c=0$, then we are done.

Let $c \ne 0$. Then

$|cA_n-cA_m|=|c||A_n-A_m| < \epsilon \iff |A_n-A_m|< \frac{\epsilon}{|c|}$.

Can you procced ?

• Does this not prove its a cauchy sequence, doesn't it satisfy the definition of a cauchy sequence? – Sam Heslop Nov 3 '17 at 12:45
• If (A_n) is Cauchy and $\epsilon >0$, then there is $N$ such that $|A_n-A_m|< \frac{\epsilon}{|c|}$ for all $n,m >N$. Hence we have $|cA_n-cA_m|< \epsilon$ for all $n,m >N$, which shows that $(cA_n)$ is Cauchy. – Fred Nov 3 '17 at 12:48
• Thank you very much been struggling to get my head around this definition – Sam Heslop Nov 3 '17 at 12:49