3
$\begingroup$

I was trying to solve a question of an entrance exam. I am having trouble in the following problem. Please help me.

For positive real numbers $a_1, a_2, \ldots, a_p$ find the value of $$\displaystyle \lim_{n\to\ \infty}\sqrt[n]{\frac{\displaystyle \sum_{i=1}^p a_i^n}{p}}$$

What I have done so far:

From AM-GM inequality $\frac{\displaystyle \sum_{i=1}^p a_i^n}{p} \ge \sqrt[p]{\displaystyle \prod_{i=1}^p a_i^n} = \sqrt[\frac{n}{p}]{\displaystyle \prod_{i=1}^p a_i}$

So $\sqrt[n]{\frac{\displaystyle \sum_{i=1}^p a_i^n}{p}} \ge \sqrt[p]{\displaystyle \prod_{i=1}^p a_i}$

But then I can not find any way to proceed further. It will be very helpful for me any one provide me some help. I apologise for not showing much effort but I am really stuck. Please help me. Thnx in advance.

$\endgroup$
  • $\begingroup$ If we help you, will that be cheating on this entrance exam? $\endgroup$ – vadim123 Apr 29 '15 at 13:14
  • $\begingroup$ @vadim123 sorry sir. But this is a question of previous year papers of the exam. I am trying to practice for the cuurent year. $\endgroup$ – usermath Apr 29 '15 at 13:17
3
$\begingroup$

Set the vector $a=(a_1, a_2, \ldots, a_p)$. Then $\|a\|_n=\sqrt[n]{\sum_{i=1}^p a_i^n}$, and the expression you have is $p^{-1/n}\|a\|_n$. Taking $n\to \infty$ gives $p^0\|a\|_\infty=\|a\|_\infty$. This is the $\infty$-norm, i.e. $$\|a\|_\infty=\max\{|a_1|,|a_2|,\ldots,|a_p|\}$$

[Note: this is a standard result in the theory of $p$-norms, see e.g. here].

$\endgroup$
  • $\begingroup$ My pleasure, glad to help. $\endgroup$ – vadim123 Apr 29 '15 at 17:09

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.