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I need to prove/disprove the following statement:

If $a_1, \dots, a_k$ are positive real numbers then

$$\lim_{n \to \infty} \sqrt[n]{\frac{a_1^n+\dots+a_k^n}{k}}=\max\{a_1,\dots,a_k\}$$

First, I'm almost certain that the statement is true just by plugging many different numbers. I tried proving it in many ways:

  • By induction on $k$
  • By a direct calculation (after some manipulations on the root)
  • By finding $N \in \mathbb{N}$ such that for every $n>N$ $$\left|\lim_{n \to \infty} \sqrt[n]{\frac{a_1^n+\dots+a_k^n}{k}}-\max\{a_1,\dots,a_k\}\right|<\varepsilon$$ (for a given $\varepsilon>0$) (Then it will satisfy the definition of a limit of a sequence)

But I failed. Any suggestions?

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An idea: suppose WLOG that $\;a_1=\max\{a_1,...,a_k\}\;$ :

$$?\xleftarrow[n\to\infty]{}\frac{a_1}{\sqrt[n]k}\le\sqrt[n]\frac{a_1^n+\ldots a_k^n}k\le\sqrt[n]\frac{ka_1^n}k=\ldots$$

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    $\begingroup$ So because $\lim_{n \to \infty} \sqrt[n]{a_1^n/k}=a_1$ and $\lim_{n \to \infty} \sqrt[n]{k a_1^n/k}=a_1$ using the "squeeze theorem" we show that the limit of the middle expression is $a_1$ as well? Brilliant. $\endgroup$ – user217919 Mar 28 '15 at 13:27
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hint:

suppose your $a_j$ are the components of a $k$-vector $a$ and $$ f_n(a)= \sqrt[n]{\frac{a_1^n+\dots+a_k^n}{k}} $$ then for any positive real $\lambda$ we have: $$ f_n(\lambda a) = \lambda f_n(a) $$ suppose, initially, there is a unique largest $a_M$ and that $a_N$ is the next largest in order of magnitude, so that $a_M - a_N \gt 0$. choose $\lambda=\frac2{a_M+a_N}$ then $\lambda a_M \gt 1$ and $$ \sum_{j=1}^k(\lambda a_j)^n \ge \lambda^na_M^n + (k-1)(\frac{a_M-a_N}{a_M+a_N})^n $$ you should be able to use this estimate to make further progress with your problem

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