## University Calculus: Early Transcendentals (3rd Edition)

$a_1=a_2=a_3=a_4=\frac{1}{2}$
We know that $a_n=\frac{2^n}{2^{n+1}}=\frac{2^n}{2\cdot2^n}=\frac{1}{2}$. Hence, here: $a_1=a_2=a_3=a_4=\frac{1}{2}$