Expectation of Sum of Gamma over Product of Inverse-Gamma Let $X_1, X_2, \cdots, X_n \sim Gamma(\alpha, \beta)$. How do we compute $E\left(\cfrac{\sum_1^n X_i}{(\prod_1^n X_i)^{1/n}}\right)$ ?
I am stuck on how to compute this expectation. I know that $\cfrac{1}{X_i}$ follows inverse-gamma but how to we handle $\cfrac{1}{(X_i)^{1/n}}$ and how do we separate the sum and the product ?
 A: This is rather straightforward (when the $X_i$'s are independent):
\begin{align*}\mathbb{E}\left(\cfrac{\sum_{i=1}^n X_i}{(\prod_{j=1}^n X_j)^{1/n}}\right) &= \sum_{i=1}^n  \mathbb{E}\left(\cfrac{ X_i}{(\prod_{j=1}^n X_j)^{1/n}}\right)\\
&= \sum_{i=1}^n  \mathbb{E}[X_i^{1-1/n}]\times \mathbb{E}\left(\cfrac{1}{(\prod_{j\ne i} X_j)^{1/n}}\right)\\
&= \sum_{i=1}^n  \mathbb{E}[X_i^{1-1/n}]\times \prod_{j\ne i}\mathbb{E}\left[X_j^{-1/n}\right]\\
&= n\mathbb{E}[X_1^{1-1/n}]\mathbb{E}\left[X_1^{-1/n}\right]^{n-1}\\
&= n\times \beta^{1/n-1}\dfrac{\Gamma(\alpha+1-1/n)}{\Gamma(\alpha)}\times\left[\beta^{1/n}\dfrac{\Gamma(\alpha-1/n)}{\Gamma(\alpha)}\right]^{n-1}\\
&=n\times \beta^{1/n-1+(n-1)/n}\times\dfrac{\Gamma(\alpha+1-1/n)\Gamma(\alpha-1/n)^{n-1}}{\Gamma(\alpha)^n}\\
&=n\times \dfrac{(\alpha-1/n)\Gamma(\alpha-1/n)^{n}}{\Gamma(\alpha)^n}\end{align*}
If considering the expectation of the ratio of the arithmetic mean to the geometric mean, i.e., when divding the above by $n$, one obtains a ratio that converges to $1$ with $\alpha\to\infty$

