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 $x_i$.


$\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 $\log z$.

| cite | improve this answer | |
  • $\begingroup$ We can also show that if $(x_i)_{i\in \mathbb N}$ is a bounded sequence with $z=\sup_i| x_i|$ then $z=\lim_{k\to \infty}2^{-k}\log \sqrt { \sum_{i\in \mathbb N} x_i^{2^{k+1}} }.$ $\endgroup$ – DanielWainfleet Apr 2 '17 at 6:03
  • $\begingroup$ 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. $\endgroup$ – marty cohen Apr 2 '17 at 12:48
  • $\begingroup$ 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 $\lim_{n\to \infty }(\int_0^1[f(x)]^n\;dx)^{1/n}=\max f(x).$ $\endgroup$ – DanielWainfleet Apr 3 '17 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.