# How to compute the $\liminf$ of the given series.

How do I compute $$\liminf_{k \to \infty}\frac{\log \sqrt{(x_1)^{2^{k+1}}+(x_2)^{2^{k+1}}+(x_3)^{2^{k+1}}+(x_4)^{2^{k+1}}+(x_5)^{2^{k+1}}+(x_6)^{2^{k+1}}}}{2^k},$$ where $x_i$ are arbitrary real numbers?

I can only think of L'Hospital's law, but it seems hard to compute.

I will assume that all the $x_i$ are positive, since they are all raised to an even power.

Let $z$ be the largest of all the $x_i$.

Then

$\begin{array}\\ \frac{\log \sqrt{\sum_{i=1}^n(x_i)^{2^{k+1}}}}{2^k} &=\frac{\log \sqrt{z^{2^{k+1}}\sum_{i=1}^n(x_i/z)^{2^{k+1}}}}{2^k}\\ &=\frac{\log z^{2^k}\sqrt{\sum_{i=1}^n(x_i/z)^{2^{k+1}}}}{2^k}\\ &=\frac{\log z^{2^k}+\log\sqrt{\sum_{i=1}^n(x_i/z)^{2^{k+1}}}}{2^k}\\ &=\frac{2^k\log z+\log\sqrt{\sum_{i=1}^n(x_i/z)^{2^{k+1}}}}{2^k}\\ &=\log z+\frac{\log\sqrt{\sum_{i=1}^n(x_i/z)^{2^{k+1}}}}{2^k}\\ \end{array}$

Since each $x_i/z \le 1$, $\sum_{i=1}^n(x_i/z)^{2^{k+1}} \le n$, so $|\frac{\log \sqrt{\sum_{i=1}^n(x_i)^{2^{k+1}}}}{2^k}-\log z| \le \frac{\log\sqrt{n}}{2^k} \to 0$ for large $k$.

So the limit, not just the lim inf, is $\log z$.

• We can also show that if $(x_i)_{i\in \mathbb N}$ is a bounded sequence with $z=\sup_i| x_i|$ then $z=\lim_{k\to \infty}2^{-k}\log \sqrt { \sum_{i\in \mathbb N} x_i^{2^{k+1}} }.$ – DanielWainfleet Apr 2 '17 at 6:03
• You have to be careful because now it is a double limit. For example, suppose the sequence is constant. Also, that should be log z, not just z. – marty cohen Apr 2 '17 at 12:48
• Yes it should be $\log z$ in my comment. Another one like it is: If $f:[0,1]\to \mathbb R$ is continuous and non-negative then $\lim_{n\to \infty }(\int_0^1[f(x)]^n\;dx)^{1/n}=\max f(x).$ – DanielWainfleet Apr 3 '17 at 18:29