How find this nice limit

Question

It is well kown this following $$\lim_{x\to+\infty}\left(\dfrac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)^x=\sqrt{ab}(a,b>0)$$ and also kown this general $$\lim_{x\to+\infty}\left(\dfrac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+a^{\frac{1}{x}}_{n}}{n}\right)^x=\sqrt[n]{a_{1}a_{2}\cdots a_{n}},(a_{i}>0,i=1,2,\cdots,n)$$

and some hours ago,I found this nice and Hard limit

$$\lim_{x\to+\infty}x\left[\left(\dfrac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)^x-\sqrt{ab}\right]=\dfrac{\sqrt{ab}}{8}\left(\ln{a}-\ln{b}\right)^2$$

my proof

$$a^{\frac{1}{x}}+b^{\frac{1}{x}}=e^{\frac{1}{x}\ln{a}}+e^{\frac{1}{x}\ln{b}}=1+\dfrac{1}{x}\ln{a}+\dfrac{1}{2x^2}\ln^2{a}+1+\dfrac{1}{x}\ln{b}+\dfrac{1}{2x^2}\ln^2{b}+o(1/x^2)$$

then

\begin{align*} \left(\dfrac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)^x&=\left[1+\dfrac{1}{2x}\ln{ab}+\dfrac{1}{4x^2}\left(\ln^2{a}+\ln^2{b}\right)+o(\frac{1}{x^2})\right]^x\\ &\approx e^{x\ln{\left(1+\dfrac{1}{2x}\ln{ab}+\dfrac{1}{4x^2}\left(\ln^2{a}+\ln^2{b}\right)\right)}}\\ &\approx e^{x\left(\dfrac{1}{2x}\ln{ab}+\dfrac{1}{4x^2}\left(\ln^2{a}+\ln^2{b}\right)\right)}\\ &=\cdots\cdots\\ &\approx \sqrt{ab}\left(1+\dfrac{1}{8x}\left(2\ln^2{a}+2\ln^2{b}-\ln^2{(ab)}\right)\right) \end{align*} so

$$\lim_{x\to+\infty}x\left[\left(\dfrac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)^x-\sqrt{ab}\right]=\dfrac{\sqrt{ab}}{8}\left(\ln{a}-\ln{b}\right)^2$$

My question:

$$\lim_{x\to+\infty}x\left[\left(\dfrac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+a^{\frac{1}{x}}_{n}}{n}\right)^x-\sqrt{a_{1}a_{2}\cdots a_{n}}\right]=?$$

Answer 1

We can compute

$$\lim_{x\to +\infty} x\left[\left(\frac{a_1^{1/x} + a_2^{1/x} + \dotsb + a_n^{1/x}}{n}\right)^x - \sqrt[n]{a_1a_2\dotsb a_n}\right]$$

in the same way. For brevity, write $\lambda_\nu = \log a_\nu$, then we have

$$\begin{align} \frac1n\sum_{\nu=1}^n a_\nu^{1/x} &= \frac1n\sum_{\nu=1}^n \exp\left(\frac1x\lambda_\nu\right)\\ &= \frac1n\sum_{\nu=1}^n \left(1 + \frac{\lambda_\nu}{x} + \frac{\lambda_\nu^2}{2x^2} + O(x^{-3})\right)\\ &= 1 + \frac{1}{nx}\sum_{\nu=1}^n\lambda_\nu + \frac{1}{2nx^2}\sum_{\nu=1}^n\lambda_\nu^2 + O(x^{-3}). \end{align}$$

Write $\mu := \frac1n\sum\limits_{\nu=1}^n\lambda_\nu$ and $s = \frac1n\sum\limits_{\nu=1}^n \lambda_\nu^2$, so we have

$$A := \frac1n\sum_{\nu=1}^n a_\nu^{1/x} = 1 + \frac{\mu}{x} + \frac{s}{2x^2} + O(x^{-3}).$$

Thus

$$\begin{align} \log A &= \frac{\mu}{x} + \frac{1}{2x^2}\left(s - \mu^2\right) + O(x^{-3}),\\ x\log A &= \mu + \frac{1}{2x} (s-\mu^2) + O(x^{-2}),\\ A^x - \sqrt[n]{a_1a_2\dotsb a_n} &= \exp \left(x\log A\right) - e^\mu\\ &= e^\mu\left(\exp (x\log A - \mu) - 1\right)\\ &= e^\mu\left(\exp \left(\frac{s-\mu^2}{2x} + O(x^{-2}) \right)-1\right)\\ &= e^\mu\left(\frac{s-\mu^2}{2x} + O(x^{-2})\right)\\ x\left(A^x - e^\mu\right) &= e^\mu\left( \frac{s-\mu^2}{2} + O(x^{-1})\right), \end{align}$$

whence

$$\begin{align} \lim_{x\to +\infty}&\; x\left[\left(\frac{a_1^{1/x} + a_2^{1/x} + \dotsb + a_n^{1/x}}{n}\right)^x - \sqrt[n]{a_1a_2\dotsb a_n}\right]\\ &= e^\mu\frac{s-\mu^2}{2}\\ &= \sqrt[n]{a_1a_2\dotsb a_n}\frac{n\sum\limits_{\nu=1}^n \log^2 a_\nu - \left(\sum\limits_{\nu=1}^n \log a_\nu\right)^2}{2n^2}. \end{align}$$