Just asking...

[Gottfried]

Thats interesting! Can you give a proof for n=0, i.e that
\( \lim_{n\to 0} \sqrt[n]{\frac{a_1^n+\dots+a_m^n}{m}} = \sqrt[m]{a_1 \dots a_m} \)?

Erm... no. Hasn't this already been proven somewhere?

Yes, but knowing an believing are two different things

When I considered that formula I realized that I never learned the logarithm formula during studying:
\( \log(x) = \lim_{n\to\infty} (\sqrt[n]{x} - 1) n \)
This is just what you get when you invert the famous Euler formula \( e^x = \lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n} \).

Knowing that formula, we can do much more with the limit
\( \lim_{\eps\to 0} \left(\frac{1^\eps + a^\eps}{2}\right)^{1/\epsilon} =\lim_{n\to\infty} \left(\frac{1+\sqrt[n]{a}}{2}\right)^n = \lim_{n\to\infty} \left(\frac{2+\sqrt[n]{a}-1}{2}\right)^n = \lim_{n\to\infty} \left(1+\frac{(\sqrt[n]{a}-1)n}{2n}\right)^n = e^{\ln(x)/2} = \sqrt[2]{x} \)

This is just a particular case with \( m=2 \), \( a_1=a \) and \( a_2=1 \) in the formula \( \lim_{\epsilon\to 0} \sqrt[\epsilon]{\frac{a_1^\epsilon +\dots+ a_m^\epsilon}{m}}=\sqrt[m]{a_1\dots a_m} \).

with the above base idea it is then no more difficult to prove the general case.
This formula is really amazing

True! I tried to prove it, too, but the without your first element I stuck.
Amazing!

Gottfried