Answer
$a_1 = \frac{(-1)^1}{2^1} = \frac{-1}{2}$
$a_2 = \frac{(-1)^2}{2^2} = \frac{1}{4}$
$a_3 = \frac{(-1)^3}{2^3} = \frac{-1}{8}$
$a_4 = \frac{(-1)^4}{2^4} = \frac{1}{16}$
Converges. The limit of the sequence seems to be $0$.
Work Step by Step
$a_n = \frac{(-1)^n}{2^n}$
$a_1 = \frac{(-1)^1}{2^1} = \frac{-1}{2}$
$a_2 = \frac{(-1)^2}{2^2} = \frac{1}{4}$
$a_3 = \frac{(-1)^3}{2^3} = \frac{-1}{8}$
$a_4 = \frac{(-1)^4}{2^4} = \frac{1}{16}$
Converges and approaches $0$. Although the signs of the terms alternate in the sequence, the absolute value of the terms is decreasing and seems to approach $0$.
