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 \(\displaystyle \lim_{\substack {a_0 \to 0 \\ a_1 \to \infty}} \text{E}\left[\frac{\text{AM}}{\text{GM}}\right]\).

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 \(X.\)