# Geometric mean as limit to power means.

This is an special case of my previous problem, however for some reason I have never been sure of the final step.

Let $a_i$ ($i=1,2,\ldots,n$) be a set of positive values, and $h,k$ real numbers (different than $0$).

It can be proved by Jensen's inequality that if $h\le k$ then $$\sqrt[h]{\frac{a_1^h+a_2^h+\cdots+a_n^h}{n}}\le\sqrt[k]{\frac{a_1^k+a_2^k+\cdots+a_n^k}{n}}.$$

It is also possible to demonstrate that if $k>0$ then $$\sqrt[k]{\frac{a_1^k+a_2^k+\cdots+a_n^k}{n}}\ge\sqrt[n]{a_1\cdot a_2\cdots a_n}.$$ Likewise if $k<0$ $$\sqrt[k]{\frac{a_1^k+a_2^k+\cdots+a_n^k}{n}}\le\sqrt[n]{a_1\cdot a_2\cdots a_n}.$$

If we see this as a function of $k$: $$f(k)=\left(\frac1n\sum_{i=1}^na_i^k\right)^{\frac1k}$$

Then $f$ is increasing and continuous for $k>0$ and $f$ is increasing and continuous for $k<0$, however it is not defined for $k=0$.

From the previous considerations seems logical that $$\lim_{k\to0}f(k)=\left(\prod_{i=1}^na_i\right)^{\frac1n}.$$ However, “it seems logical” is barely a proof.

My attempts. Case $n>2$ can be easily reduced by a combination of induction on $m$ for $n=2^m$ and reverse induction (if it is true for $n=m+1$ prove that it is true for $n=m$) [or by direct induction if we can generalize a weighted version.] So it should be enough to prove for $n=2$.

The we must prove that $$\lim_{k\to0}\left(\frac{a^k+b^k}2\right)^{\frac1k}=\sqrt{ab}.$$

I will simplify further. If I escalate by a factor of $\alpha$, then $$\left(\frac{(\alpha a)^k+(\alpha b)^k}{2}\right)^{\frac1k}=\left(\frac{\alpha^ka^k+\alpha^kb^k}{2}\right)^{\frac1k}=\left(\alpha^k\frac{a^k+b^k}{2}\right)^{\frac1k}=\alpha\left(\frac{a^k+b^k}{2}\right)^{\frac1k}$$ likewise $\sqrt{(\alpha a)(\alpha b)}=\alpha\sqrt{ab}$, so I loose no generality if I make $a=x$ and $b=1/x$ for any positive real $x$.

So my problem is reduced to prove: $$\lim_{k\to0}\left(\frac{x^k+x^{-k}}2\right)^{\frac1k}=1.$$ (note: for general $a,b$ take $\alpha=\sqrt{ab}$ and $x=\sqrt{\frac 1b}$.)

Now, let $y=x^k$ (so $k=\log_xy$), as $k\to0$ then $y\to1$. Then \begin{align} \lim_{k\to0}\left(\frac{x^k+x^{-k}}2\right)^{\frac1k}=&\lim_{y\to1}\left(\frac{y^2+1}{2y}\right)^{\frac1{\log_xy}}\\ &=\exp\lim_{y\to1}\frac1{\log_xy}\ln\frac{y^2+1}{2y}\\ &=\exp\lim_{y\to1}\frac{\ln x}{\ln y}\ln\frac{y^2+1}{2y}\\ &=\exp\ln x\lim_{y\to1}\frac{\ln\frac{y^2+1}{2y}}{\ln y}\\ \end{align} Now, using L'hopital: $$\frac d{dy}\ln\frac{y^2+1}{2y}=\frac d{dy}\ln(y^2+1)-\ln2-\ln y = \frac{2y}{y^2+1}-\frac1y$$ which evaluates as $0$ when $y=1$, while $\frac d{dy}\ln y=\frac1y$ which is non-zero.

So: $$\lim_{k\to0}\left(\frac{x^k+x^{-k}}2\right)^{\frac1k}=\lim_{y\to1}\left(\frac{y^2+1}{2y}\right)^{\frac1{\log_xy}}=\exp\ln x\lim_{y\to1}\frac{\ln\frac{y^2+1}{2y}}{\ln y}=x^0=1.$$

My problem: Well, this proof seem too long and I am not completely convinced that I can change $\lim_{k\to0}$ into $\lim_{y\to1}$.

Is there a simpler way to prove $$\lim_{k\to0}\left(\frac{a^k+b^k}2\right)^{\frac1k}=\sqrt{ab}?$$

• Commented Oct 25, 2013 at 20:30
• Thank you @Guest. I knew it should be way simpler. Commented Oct 25, 2013 at 20:39