Find $\lim_{n\to\ \infty}\sqrt[n]{\frac{\sum_{i=1}^p a_i^n}{p}}$

Question

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.

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

score 3 · Accepted Answer · edited Apr 13 '17 at 12:20

3

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].

edited Apr 13 '17 at 12:20

Community

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answered Apr 29 '15 at 13:21

vadim123

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My pleasure, glad to help. – vadim123 Apr 29 '15 at 17:09

Find $\lim_{n\to\ \infty}\sqrt[n]{\frac{\sum_{i=1}^p a_i^n}{p}}$

1 Answers1