University Calculus: Early Transcendentals (3rd Edition)

$\sum\limits_{k=1}^5(-1)^{k+1}\cdot\frac{1}{k}$
As we can see, the fraction's denominator is always growing by one, and the fraction's sign is always changing, therefore the formula is: $\sum\limits_{k=1}^5(-1)^{k+1}\cdot\frac{1}{k}$