Answer
Converges to $1$
Work Step by Step
$a_n = 2^nsin(2^{-n})$
$a_1 = 0.958851$
$a_2 = 0.989616$
$a_3 = 0.997398$
$a_4 = 0.999349$
$a_5 = 0.999837$
$a_6 = 0.999959$
$a_7 = 0.99999$
$a_8 = 0.999997$
$a_9 = 0.999999$
$a_{10} = 1.0$
The terms of the sequence seems to get closer and closer to $1$.