Answer
$\displaystyle \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots+\frac{13}{13+1}=\sum_{k=1}^{13}\frac{k}{k+1}$
Work Step by Step
$\displaystyle \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots+\frac{13}{13+1}=$
There are 13 terms, as the numerators indicate.
Count them with index k: $\displaystyle \sum_{k=1}^{13}$(...)
The first term (when k=1) is $\displaystyle \frac{1}{2}$,
The second term (when k=2) is $\displaystyle \frac{2}{3}$,
The kth is $\displaystyle \frac{k}{k+1}$ so
$\displaystyle \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots+\frac{13}{13+1}=\sum_{k=1}^{13}\frac{k}{k+1}$
