It is easy to prove that
$\lim\_{p\rightarrow 1} m\_p = (x\_1 + \cdots + x\_n)/n$. The following fact about the derivative of $m\_p$ with respect to $p$ is also elementary:
$$m'\_p =\frac{dm\_p}{dp}
=\frac{1}{p \log p}\cdot\Big[\frac{x\_1p^{x\_1}+\cdots+x\_np^{x\_n}}{p\_1^{x\_1}+\cdots+p\_n^{x\_n}}-m\_p\Big].$$
My interest in this is to create an alternative to the power mean, called the *exponential mean*: see [here](https://math.stackexchange.com/questions/3800733/inequalities-for-generalized-means) and [here](https://www.datasciencecentral.com/profiles/blogs/alternative-to-the-arithmetic-geometric-and-harmonic-means). The limit I am interested in is $\lim\_{p\rightarrow 1} m'\_p$. Using WolframAlpha, I computed the limit for $n=2,3,4,5$ (see [here](https://www.wolframalpha.com/input/?i=lim%20%28%28x%20p%5Ex%20%2B%20y%20p%5Ey%20%2B%20z%20p%5Ez%20%2B%20w%20p%5Ew%20%2B%20u%20p%5Eu%29%2F%28p%5Ex%20%2B%20p%5Ey%20%2Bp%5Ez%20%2B%20p%5Ew%20%2B%20p%5Eu%29%20-%20%28log%28p%5Ex%20%2B%20p%5Ey%20%2Bp%5Ez%20%2B%20p%5Ew%20%2Bp%5Eu%29-log%285%29%29%2F%28log%20p%29%29%2F%28log%20p%29%20as%20p-%3E1)) and the following remarkable pattern emerges:
$$\lim\_{p\rightarrow 1} m'\_p=\frac{1}{2n^2}\sum\_{1\leq i<j\leq n}(x\_i-x\_j)^2.$$
How do you go about formally proving this fact? It does not sound elementary to me. Also, it sounds like $m\_p$ is a strictly increasing function of $p$ (its derivative beeing positive everywhere, with $m'\_0 =+\infty$ and $m'\_\infty =0$) unless all the $x\_i$'s are identical.


**Update**


In short, $m\_1$ is the arithmetic mean and $m'\_1$ is half the empirical variance of $x\_1,\cdots,x\_n$. I tried to see if such simple formulas existed for the power mean $M\_p$, but I could not find anything interesting other than the well known fact that $M\_1=m\_1$ is the arithmetic mean. It would be interesting to see how the second and third derivatives of $m\_p$ at $p=1$ are linked to the higher empirical moments of $x\_1,\cdots,x\_n$.



Remarkable limit involving $m_p=\log_p(p^{x_1} + \cdots + p^{x_n})-\log_p(n)$
$\newcommand\bar\overline$
Letting $t:=\ln p$, we see that the limit in question is the limit of
$$d(t):=\frac1t\Big(\sum\_1^n x\_j e^{tx\_j}\Big/\sum\_1^n e^{tx\_j}-m\_{e^t}\Big)$$
as $t\to0$.
Next, letting
$\bar x:=\frac1n\,\sum\_1^n x\_j$, $\bar{x^2}:=\frac1n\,\sum\_1^n x\_j^2$, and $s^2=\bar{x^2}-\bar x^2$, we have
$$\sum\_1^n x\_j e^{tx\_j}=\sum\_1^n x\_j (1+tx\_j+o(t))
=n(\bar x+t\bar{x^2})+o(t),$$
$$\sum\_1^n e^{tx\_j}=\sum\_1^n (1+tx\_j+o(t))
=n(1+t\bar x)+o(t),$$
$$m\_{e^t}=\log\_{e^t}\Big(\frac1n\,\sum\_1^n e^{tx\_j}\Big) \\
=\log\_{e^t}(1+t\bar x+t^2\bar{x^2}/2+o(t^2)) \\
=\tfrac1t\,\ln(1+t\bar x+t^2\bar{x^2}/2+o(t^2)) \\
=\bar x+ts^2/2+o(t).$$
So,
$$d(t)=\frac1t\Big(\frac{\bar x+t\bar{x^2}}{1+t\bar x}+o(t)-\bar x-ts^2/2\Big) \\
=\frac1t\Big((\bar x+t\bar{x^2})(1-t\bar x)+o(t)-\bar x-ts^2/2\Big) \\ 
=s^2/2+o(1). $$
So, the limit in question is
$$s^2/2
=\frac1{4n^2}\sum\_{1\le i,j\le n}(x\_i-x\_j)^2 \\ 
=\frac1{4n^2}\sum\_{1\le i,j\le n,\ i\ne j}(x\_i-x\_j)^2 \\ 
=\frac1{2n^2}\sum\_{1\le i<j\le n}(x\_i-x\_j)^2,$$
as conjectured.




---


*Details on the first equality in the last three-line display*: The left-hand side of that equality is $\frac12\,Var\,X$, where $X$ is any random variable whose distribution is $\frac1n\,\sum\_1^n\delta\_{x\_j}$, where $\delta\_a$ is the Dirac probability measure at point $a$. The right-hand side of that equality is $$\frac14\,E(X-X')^2=\frac14\,Var(X-X')=\frac12\,Var\,X,$$
where $X'$ is an independent copy of $X$.


Of course, that equality can also be checked by straightforward algebraic calculations.


