# Thread:

1. ## Power mean

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

2. 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

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

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

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

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

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