## Precalculus (10th Edition)

$\sum_{k=1}^{13} (-1)^{k+1} (\frac{1}{k})$
We can see that this is the sum of the $13$ fractions in which the numerator is $1$, and in which the first denominator is $1$ and increases by $1$ in each term. Also, the signs are alternating, starting with $+$. Hence, the summation formula is: $\sum_{k=1}^{13} (-1)^{k+1} (\frac{1}{k})$