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Let $a_1,\ldots,a_p$ be positive real numbers. Find the limit of $$\left(\frac{a_1^n+\cdots+a_p^n}{p}\right)^{1/n}$$

My attempt:

I applied Hölder's and have obtained that this term is bounded below by $\frac{a_1+\cdots+a_p}{p}$ and since $a_1^p+\cdots+a_n^p \le (a_1+\cdots+a_p)^n$ is valid, applying log of limits technique, I get that it is bounded above by $a_1+\cdots+a_p$. However, I haven't obtained an actual limit. In fact I don't know the actual answer. I think it could be the lower bound I obtained, because that answer is validated for certain examples, like, by putting all values of $a_i$'s as a constant $k$, but I am not able to get a suitable idea to conclude that. Any help?

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  • $\begingroup$ You mean limit as $n\to\infty$ or as $p\to\infty$ or something else? ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 25 '15 at 19:51
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Let $a=\max(a_1, a_2, \dots, a_p)$. Then the limit is $$\lim_{n\to\infty} a\left( \frac{\sum_i (a_i/a)^n}{p}\right)^{1/n}=a$$

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Note that $$ \left(\frac1p\sum_{i=1}^pa_i^n\right)^{1/n}\le\left(\frac1p\sum_{i=1}^p\max_k(a_k)^n\right)^{1/n}=\max_k(a_k) $$ and $$ \left(\frac1p\sum_{i=1}^pa_i^n\right)^{1/n}\ge\left(\frac1p\max_k(a_k)^n\right)^{1/n}=\left(\frac1p\right)^{1/n}\max_k(a_k) $$ Therefore, $$ \left(\frac1p\right)^{1/n}\max_k(a_k)\le\left(\frac1p\sum_{i=1}^pa_i^n\right)^{1/n}\le\max_k(a_k) $$ By the Squeeze Theorem, we get $$ \lim_{n\to\infty}\left(\frac1p\sum_{i=1}^pa_i^n\right)^{1/n}=\max_k(a_k) $$

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