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

$\sum\limits_{k=1}^5(-1)^{k}\cdot\frac{k}{5}$
The denominators of the fractions are fives, the sign is always changing, and the numerator always grows by one, therefore the formula is: $\sum\limits_{k=1}^5(-1)^{k}\cdot\frac{k}{5}$