1
$\begingroup$

Let $\{x_n\}$ and $\{y_n\}$ be seqeunces such that $\lim y_n = 0$. Suppose that for all $k$ in $N$ and all $m \ge k$ we have $|x_m - x_k| \le y_k$. Prove $\{x_n\}$ is cauchy.

Attempt:

Since $\lim y_n = 0$, there exists $N$ in natural numbers such that for all $n \ge N$, $ |y_n| < \epsilon$.

Then for all $m,k \ge N$, $0 \le |x_m - x_k| \le y_k = |y_k| < \epsilon$.

Therefore ${x_n}$ is cauchy.

Is this a good proof? Also, is there some site where I can see steps by step for different proofs? I've been constantly coming here to make sure my proofs are right. I'm never sure if they're right or not.

$\endgroup$
  • $\begingroup$ You are alternating between $n's$ and $k's$ i.e. you say that $|y_n| <\epsilon$ on one line and $y_k < \epsilon$ on the next. Otherwise, that is the general approach I would take. $\endgroup$ – Doug M Mar 23 '17 at 4:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.