Solution:
Assume for contradiction that it was rational, and let $q$ denote its value. Squaring, we find that
$$q^{2} = (n+1) + 2 \sqrt{n} \cdot \sqrt{n+1} + n$$
so
$$\frac{q^{2} - (2n+1)}{2} = \sqrt{n(n+1)}$$
The left-hand side is also rational, so we conclude the quantity $n(n+1)$ is the square of a rational number. Actually, since it is an integer, it follows that $n(n+1)$ must be a perfect square (the square of an integer). However, $n^{2} < n(n+1) < (n+1)^{2}$, so $n(n+1)$ lies strictly between two consecutive perfect squares, which is a contradiction.