Define the following family of one parameter sequences:
$$a_0=x,~~~b_0=y$$
$$a_{n+1}=\sqrt{a_n \sqrt[p]{\frac{a_n^p+b_n^p}{2}}},~~~b_{n+1}=\sqrt{b_n \sqrt[p]{\frac{a_n^p+b_n^p}{2}}}$$
I conjecture that this family of sequences has the limit:
$$L_p(x,y)=\left(\frac{x^{2p}-y^{2p}}{2p(\ln x-\ln y)} \right)^{\dfrac{1}{2p}}$$
The proof:
$$\ln a_n-\ln b_n=\frac{\ln x-\ln y}{2^n}=\frac{\delta_1}{2^n} \tag{1}$$
$$a_n^{2p}-b_n^{2p}=(a_{n-1}^p-b_{n-1}^p)\frac{a_{n-1}^p+b_{n-1}^p}{2}=\frac{x^{2p}-y^{2p}}{2^n}=\frac{\delta_2}{2^n} \tag{2}$$
From $(2)$:
$$2p \ln a_n=2p \ln b_n+\ln \left(1+\frac{\delta_2}{2^nb_n^{2p}} \right)$$
From $(1)$:
$$2p \ln a_n=2p \ln b_n+\frac{2p\delta_1}{2^n}$$
$$\ln \left(1+\frac{\delta_2}{2^nb_n^{2p}} \right)=\frac{2p\delta_1}{2^n} \tag{3}$$
Assuming $2^nb_n^{2p} \to \infty$ at $n \to \infty$ we expand the logarithm, considering only the first term:
$$\frac{\delta_2}{2^nb_n^{2p}}+O\left(\frac{1}{2^{2n}} \right)=\frac{2p\delta_1}{2^n}$$
$$\lim_{n \to \infty} b_n=\left(\frac{\delta_2}{2p \delta_1} \right)^{\dfrac{1}{2p}}=\left(\frac{x^{2p}-y^{2p}}{2p(\ln x-\ln y)} \right)^{\dfrac{1}{2p}} \tag{4}$$
Is this proof correct for any $p \in \mathbb{R}$?
How would I know that $2^nb_n^{2p} \to \infty$, i.e. $b_n^{2p}$ doesn't vanish?
Does the mean defined by $(4)$ has a special name?
Also, do you know other such parametric families of sequences, related to means?
The case $p=\frac{1}{2}$ is familiar and appeared in my recent question.