For this question, I know this sequence converges, but I'm not sure how to get to the final answer. Here is what I have so far, can anyone please help me out?
Determine whether the sequence converges or diverges.
$\left\{\dfrac{1}{2^n} -3\right\}$
$\lim_{n\to \infty} \left(\dfrac{1}{2^n} -3\right) = -3$
Let $\epsilon>0$ be arbitrary.
Choose N $ \geq 0$
Suppose $n>N$
$\left|\dfrac{1}{2^n} -3 -(-3)\right| < \epsilon$
$\left|\dfrac{1}{2^n}\right| < \epsilon$
$\dfrac{1}{2^n} < \dfrac{1}{2^N}$
From here we obtain
$$\left|\frac{1}{2^n}\right| < \epsilon\iff 2^n>\frac1{\epsilon}\iff n>\log_2\frac1{\epsilon}\implies n>N=\lceil\log_2\frac1{\epsilon}\rceil$$