1. ## Power mean

From the power mean inequality $\displaystyle P_r = \left(\frac{a_{1}^{r} + \ldots + a_{n}^{r} }{n}\right)^{1/r}$ why does $\displaystyle \lim_{r \to 0} P_{r} = (a_{1}a_{2} \cdots a_{n})^{1/n}$?

2. Originally Posted by particlejohn
From the power mean inequality $\displaystyle P_r = \left(\frac{a_{1}^{r} + \ldots + a_{n}^{r} }{n}\right)^{1/r}$ why does $\displaystyle \lim_{r \to 0} P_{r} = (a_{1}a_{2} \cdots a_{n})^{1/n}$?
There's a proof in this link: Generalized mean - Wikipedia, the free encyclopedia

3. First: $\displaystyle \mathop {\lim }\limits_{r \to 0} \log \left( {P_r } \right) = \mathop {\lim }\limits_{r \to 0} \tfrac{1} {r} \cdot \log \left( {\tfrac{{\sum\nolimits_{k = 1}^n {a_k ^r } }} {n}} \right)$

Now we have: $\displaystyle \log \left( {\tfrac{{\sum\nolimits_{k = 1}^n {a_k ^r } }} {n}} \right)\mathop \sim \limits_{r \to 0} \tfrac{{\sum\nolimits_{k = 1}^n {a_k ^r } }} {n} - 1 = \tfrac{{\sum\nolimits_{k = 1}^n {\left( {a_k ^r - 1} \right)} }} {n}$ (since what's inside the logarithm tends to 1)

So: $\displaystyle \mathop {\lim }\limits_{r \to 0} \log \left( {P_r } \right) = \mathop {\lim }\limits_{r \to 0} \tfrac{1} {r} \cdot \tfrac{{\sum\nolimits_{k = 1}^n {\left( {a_k ^r - 1} \right)} }} {n} = \tfrac{1} {n} \cdot \sum\nolimits_{k = 1}^n {\mathop {\lim }\limits_{r \to 0} \tfrac{1} {r} \cdot \left( {a_k ^r - 1} \right)}$ (by the linearity of the limit- all of those are defined-)

$\displaystyle \mathop {\lim }\limits_{r \to 0} \tfrac{1} {r} \cdot \left( {a_k ^r - 1} \right) = \mathop {\lim }\limits_{r \to 0} \tfrac{1} {r} \cdot \left( {e^{r \cdot \log \left( {a_k } \right)} - 1} \right) = \log \left( {a_k } \right)$

So: $\displaystyle \mathop {\lim }\limits_{r \to 0} \log \left( {P_r } \right) = \tfrac{1} {n} \cdot \sum\nolimits_{k = 1}^n {\log \left( {a_k } \right)}$ and the rest follows easily by the properties of the logarithm and the continuity of the exponential function.