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Random Talk on Random Thoughts

Power Means With Infinite Exponents

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Half a month ago, I didn’t know how to find $\lim\limits_{n \to \infty} \sqrt[n]{a^n+b^n}$. With the help from others, I could show that the answer was $\max\left\{ a,b \right\}$. This inspired me to solve a question which had been in my mind since I was F.3.

Suppose that $M_n (a_1,\dots,a_k) := \sqrt[n]{\frac1k \sum\limits_{i=1}^{k} a_i^n} \quad\forall\,n \in \N$, and $a_i > 0\quad\forall\,i \in \{1, \dots,k\}$. Show that $\lim\limits_{n \to \infty} M_n (a_1,\dots,a_k) = \max\limits_{i \in \{ 1,\dots,k\}} a_i$ and $\lim\limits_{n \to -\infty} M_n = \min\limits_{i \in \{1,\dots,k\}} a_i$.


I'll use the facts that $\lim\limits_{n \to \infty} b^{1/n} = 1 \quad\forall\,b>0$. (It can be proved by dividing $b$ into $0 < b < 1$ and $b > 1$. For $b > 1$, let $b^{1/n} = 1 + \delta_n$ for some $\delta_n > 0$. It's a good exercise on the definition of limits, the binomial expansion and elementary properties of inequalities.)

Let $M := \max\limits_{i \in \{1,\dots,k\}} a_i$. Note that

\[ \frac{M}{k^{1/n}} = \sqrt[n]{\frac{M^n}{k}} \le \sqrt[n]{\frac1k \sum\limits_{i=1}^{k} a_i^n} \le \sqrt[n]{\frac{kM^n}{k}} = M. \]

Therefore, taking limit as $n \to \infty$ and applying the Squeeze Theorem, one has

\[ \lim_{n \to \infty} M_n (a_1,\dots,a_k) = M = \max\limits_{i \in \{ 1,\dots,k\}} a_i. \]

I wrote similar arguments for the case $n \to -\infty$, but after I read Wikipedia, I’ve learnt a quicker way to finish the question.

Replace $a_i$'s with $1/a_i$'s. Then

\[ \begin{aligned} \lim_{n \to \infty} \sqrt[n]{\frac1k \sum\limits_{i=1}^{k} \left( \frac{1}{a_i} \right)^n} &= \max_{i \in \{1,\ldots,k\}} \frac{1}{a_i}\\ \lim_{n \to \infty} \sqrt[n]{\frac1k \sum\limits_{i=1}^{k} a_i^{-n}} &= \frac{1}{\min\limits_{i \in \{1,\ldots,k\}} a_i} \\ \lim_{n \to \infty} \left( \frac1k \sum\limits_{i=1}^{k} a_i^{-n}\right)^{-1/n} &= \min\limits_{i \in \{1,\ldots,k\}} a_i \end{aligned} \]

Hence, $\lim\limits_{n \to -\infty} M_n (a_1,\dots,a_k) = \lim\limits_{n \to -\infty} \sqrt[n]{\frac1k \sum\limits_{i=1}^{k} a_i^n} = \min\limits_{i \in \{1,\ldots,k\}} a_i$.