Power mean

**particlejohn** · August 29th 2008, 08:37 PM

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

**mr fantastic** · August 29th 2008, 09:12 PM

Originally Posted by particlejohn

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

There's a proof in this link: Generalized mean - Wikipedia, the free encyclopedia

**PaulRS** · August 30th 2008, 05:16 AM

First: $ \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: $ \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: $ \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-)

$ \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: $ \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.

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