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For a positive real numbers $a_1, a_2,.... a_p$ what is

$ \lim_{n\to \infty}(\frac{ (a_1^n+a_2^n+....+a_p^n)}{p})^\frac{1}{n}$

Now I apply Cauchy root test on this and i evaluated $ \lim_{n\to \infty} \frac{a_{n+1}}{a_n}$ it comes out

$ \lim_{n\to \infty}$ $ \frac{(a_1^{n+1}+a_2^{n+1}+....+a_p^{n+1})}{(a_1^{n}+a_2^{n}+....+a_p^{n}})$.

then i apply L hospital rule?

Am i right?

zhw.
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gaurav
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  • Relevant: http://math.stackexchange.com/questions/326172/the-l-infty-norm-is-equal-to-the-limit-of-the-lp-norms – Clement C. May 06 '16 at 13:40
  • What is Cauchy root test and why are you looking at $a_{n+1}/a_n?$ – zhw. May 06 '16 at 16:23
  • i mean cauchy second theorem on limits accoring to which $\lim_{n\to \infty} (a_n)^\frac{1}{n}= \lim_{n\to \infty}\frac{a_{n+1}}{a_n}$ – gaurav May 07 '16 at 03:07

1 Answers1

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I'm not sure what you are doing, but it doesn't look promising to me. Here's a hint: Let $M= \max(a_1,\dots , a_p).$ Then

$$M^n \le \sum_{k=1}^{p} a_k^n \le pM^n.$$

zhw.
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