Consider this pattern where the positive, proper fractions with denominator $(n+1)$  are arranged in the $n$th row in a triangular formation. The 1st through 4th rows are shown; each row has one more entry than the previous row. What is the sum of the fractions in the 15th row?

[asy]
label("$\frac{1}{2}$",(0,0),S);
label("$\frac{1}{3}$",(-5,-5),S);
label("$\frac{2}{3}$",(5,-5),S);
label("$\frac{1}{4}$",(-10,-10),S);
label("$\frac{2}{4}$",(0,-10),S);
label("$\frac{3}{4}$",(10,-10),S);
label("$\frac{1}{5}$",(-15,-15),S);
label("$\frac{2}{5}$",(-5,-15),S);
label("$\frac{3}{5}$",(5,-15),S);
label("$\frac{4}{5}$",(15,-15),S);
dot((0,-22));
dot((0,-20));
dot((0,-24));
[/asy]
The fractions in the $n^{\text{th}}$ row are $1/(n + 1)$, $2/(n + 1)$, $\dots$, $n/(n + 1)$, so their sum is \[\frac{1 + 2 + \dots + n}{n + 1}.\]For all $n$, $1 + 2 + \dots + n = n(n + 1)/2$, so \[\frac{1 + 2 + \dots + n}{n + 1} = \frac{n}{2}.\]In particular, the sum of the fractions in the 15th row is $\boxed{\frac{15}{2}}$.