convergence of sequences $|a_{n}-L|<1/n$

Question

$a_{n}=\frac{5n}{4n-3}$ , $L=5/4$ show that for all $n>12$ , $|a_{n}-L|<1/n$.

right now, I have: $a_{n}-L=\frac{15}{4(4n-3)}<\frac{4}{4n-3}$ since $15/4=3.75<4$ then I know $\frac{4}{4n-3} > 4/4n=1/n$ since $4n-3<n$ but this is a contradiction to what I need to prove.

Answer 1

$$\dfrac{15}{16n-12}\le \frac{1}{n}$$ for all $n\ge 12.$