How much larger is the AM than the GM?

From the AM-GM inequality, we know that the arithmetic mean (AM) of a list of non-negative real numbers is always greater than or equal to the geometric mean (GM). But the inequality doesn't tell us just how much larger the AM is.

Given a list of $n$ random real numbers chosen uniformly and independently in the range $[a_0,a_1],$ where $0\le a_0<a_1,$ find $\text{E}\left[\frac{\text{AM}}{\text{GM}}\right]$ in terms of $n,a_0,a_1.$

Then find $\displaystyle \lim\limits_{\overset{a_0 \to 0}{a_1 \to \infty}} \text{E}\left[\frac{\text{AM}}{\text{GM}}\right]$.

If the formula is of the form $\large\frac{n^n}{\left(An-B\right){\left(n-1\right)}^{n-C}},$ where $A,B,C$ are positive integers, give your answer as $A+B+C.$

Note: The notation $\text{E}[X]$ is the expected value of the random variable $X.$