# How much larger is the AM than the GM?

Calculus Level 3

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 $$X.$$

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