The n-th term of this sum would be n/(n + 1) - the denominator of each terms is 1 greater than the numerator. In sigma notation, we can write
[tex]\displaystyle \frac12 + \frac23 + \frac34 + \frac45 + \frac56 = \sum_{n=1}^5 \frac{n}{n+1}[/tex]
since the first term has numerator 1 and the last term has numerator 5.
If the sum is infinite, we would instead write
[tex]\displaystyle \frac12 + \frac23 + \frac34 + \frac45 + \frac56 + \cdots = \sum_{n=1}^\infty \frac{n}{n+1}[/tex]
