Answer
$\sum_{k=1}^{13}(-1)^{k+1} \dfrac{1}{k}$
Work Step by Step
We notice that the sum of the $13$ fractions has a numerator starting with $1$ (so, we set it to $1$ in the formula) . The denominator starts from $1$ and gets increased by $1$ in each term (so, we set it to $k$ in the formula). The sign alternates at each new term as well, so we add $(-1)^{k+1}$ to the formula.
So, we determine the summation formula as: $\sum_{k=1}^{13}(-1)^{k+1} \dfrac{1}{k}$
