Proving this Cauchy Sequence

Question

For $\varepsilon = \frac{1}{100}$ find $N \in \mathbb{N}$ such that $n \geq N$ implies $$\bigg|\frac{5n}{3n+1}-\frac{5}{3}\bigg|<\frac{1}{100}$$

I know that for a sequence {$a_{n}$} of real or complex numbers is a Cauchy Sequence if for every $\varepsilon > 0$, there exists a natural number $N$ such that if $n \geq N$ and $m\geq N$ then $|a_{n}-a_{m}|<\varepsilon$.

Thank you.

Answer 1

$ \bigg|\frac{5n}{3n+1}-\frac{5}{3}\bigg|<\frac{1}{100} \quad \iff \quad\frac{5}{9n+3}<\frac{1}{100} \quad \iff \quad 9n> 497$.

Your turn !