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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.

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    $\begingroup$ It might help you to combine $${3^n+2\over3^n+1}-1$$ into a single fraction. $\endgroup$ – Gerry Myerson Aug 4 '13 at 11:38
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Hints:

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

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

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  • $\begingroup$ I have nothing more that you noted ;) Just to make a machinery approach. That's it. Of course the OP didn't want it:) $\endgroup$ – mrs Aug 4 '13 at 12:59
  • $\begingroup$ i arrived here : -1/28<=1/(3^n+1)<=1/28 and i don't know what else to do.I tried to reduce to a common denominator and i got this: −3n−1≤28≤3n+1 Can you tell me the next step? $\endgroup$ – marinaaaa Aug 4 '13 at 13:45
  • $\begingroup$ @marinaaaa, you don't need all that: the expression inside the absolute value is always positive! You can erase the absolute value and work without it, as shown above... $\endgroup$ – DonAntonio Aug 4 '13 at 14:12
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the answer is [3, infinity) because n is greater than or = 3

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