Finding $x$ from inequality: $\left | \frac{3^n + 2}{3^n + 1} - 1 \right | \le \frac{1}{28}$

Question

Find $x$ in $\mathbb{Z}$ satisfying this inequality:

$$\left | \frac{3^n + 2}{3^n + 1} - 1 \right | \le \frac{1}{28}.$$

I tried something, but I don't think it's correct.

$$-\frac{1}{28} \le \frac{3^n + 2}{3^n + 1} - 1 \le \frac{1}{28}$$

I arrived at:

$$-3^n - 1 \le 28 \le 3^n + 1.$$

I don't know if it's ok or not.

Answer 1

Hints:

$$\left|\;\frac{3^n+2}{3^n+1}-1\;\right|=\frac1{3^n+1}\le\frac1{28}\iff\ldots$$

Note that $\,3^n+1>0\;,\;\;\forall\,n\in\Bbb N\;$

Answer 2

the answer is [3, infinity) because n is greater than or = 3