Answer
converges to $-1$.
Work Step by Step
As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists.
Consider $a_n=\dfrac{1-2^n}{2^ n}$
Apply limits to both sides.
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}[\dfrac{1-2^n}{2^ n}]$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{1}{2^ n}-\lim\limits_{n \to \infty}\dfrac{2^n}{2^ n}$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{1}{2^ n}-1$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}(\dfrac{1}{2})^ n-1$
$\lim\limits_{n \to \infty}a_n=0-1=-1$
Therefore, the sequence converges to $-1$.
