Use the definition of a Cauchy sequence to prove that ($\frac{n}{n+3}$) is Cauchy.

Question

I know that a sequence is Cauchy if for every $\epsilon>0$, there exists a natural number $N$ so that for all $n,m \geq N$, we have $|a_n-a_m|\leq\epsilon$. I'm not sure where to go from here. It doesn't help that I missed the day where we talked about Cauchy sequences. I'm confused about defining the $N$ at this point. I can do $|\frac{n}{n+3} - \frac{m}{m+3}|=|\frac{3(n-m)}{(m+3)(n+3)}|$, but I get stuck there. I know we want this to be less than or equal to $\epsilon$, but I'm unsure how to simplify it to get the correct $N$.

Answer 1

Write $m \geq n \geq N$ and, in fact, $m = n + p$ (with $p \geq 0$). Then you get $\dfrac{3p}{(n + p + 3)(n + 3)} \leq \dfrac{3}{N + 3} \leq \varepsilon$ if $N \geq ...$