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

$\sum\limits_{k=1}^4\frac{1}{2^k}$
$\frac{1}{2}=\frac{1}{2^1}, \frac{1}{4}=\frac{1}{2^2},\frac{1}{8}=\frac{1}{2^3},\frac{1}{16}=\frac{1}{2^4}$, therefore the kth element is $\frac{1}{2^k}$, therefore the formula is: $\sum\limits_{k=1}^4\frac{1}{2^k}$