Find a $N \in I \ni \left|\frac{2n}{n+3} - 3\right| < \frac{1}{5}\,\forall\,n\geq N$

Question

I have a question in my book

Find an $N \in I \ni \left|\dfrac{2n}{n+3} - 3\right| < \dfrac{1}{5}\,\forall\,n\geq N$

(I believe $I$ is the set of integers.)

But I think this finding a value $N \in I$ is not possible, because if I take $s_n = \dfrac{2n}{n+3}$, we know $\lim_{n \to \infty} s_n = \lim_{n \to \infty} \dfrac{2}{1 + \tfrac{3}{n}} = 2$.

Did I go wrong somewhere, or should I simply write $N \in I$ doesn't exist? Please explain.

Answer 1

If $n = -9$, then \begin{align} \left|\dfrac{2n}{n+3} - 3\right| &= \left|\dfrac{2(-9)}{-9+3} - 3\right|\\ &= |3 - 3| = 0 < \frac{1}{5} \end{align} So $n = -9$ is an integer that works.