Why is the $0$th power mean defined to be the geometric mean?
Solution 1:
Use the fact that as $r \rightarrow 0$, for all $a>0$: $$a^r = 1 + r \log a + o(r)$$ So the sum becomes: $$\frac{1}{n}\sum_{i=1}^n a_i^r=\frac{1}{n}\sum_{i=1}^n(1 + r \log a_i + o(r))=1+r\frac{1}{n}\sum_{i=1}^n\log a_i + o(r)$$
Furthermore $\lim_{r \rightarrow 0} (1+rx + o(r))^{1/r}=e^x$.
$$\lim_{r \rightarrow 0} \left(\frac{1}{n}\sum_{i=1}^n a_i^r\right)^{1/r}=\lim_{r \rightarrow 0} \left(1+r\frac{1}{n}\sum_{i=1}^n\log a_i + o(r)\right)^{1/r}=e^{\left(\frac{1}{n}\sum_{i=1}^n\log a_i\right)}= \left( \prod_{i=1}^n a_i\right)^{1/n}.$$