Convergence of the generalized mean

Question

Given $k\in\mathbb{N}$ real numbers $x_1,\dots ,x_k>0$, consider the following sequence of means:

$a_n:=\sqrt[n]{\sum_{j=1}^kx_j^n}$, $n\in\mathbb{N}$

I know that the limit of this sequence should be $\max\{x_1,\dots,x_k\}$. How can I prove this?

Answer 1

Let's assume first that $x_1>x_i$ for $i>1$.

Then,

$$\sum_{j=1}^k x_j^n = x_1\left(1 + \sum_{j=2}^k\left(\frac{x_j}{x_i}\right)^n\right)$$

and $0<\frac{x_j}{x_1}<1$. It should be easy to see what happens in the limit.

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Now you should also tackle the problem if there is more than one maximum element. In that case, you can assume without loss of generality that $x_1=x_2=\dots x_m>x_i$ for $i>m$, and rewrite the expression similarly. You should only get an extra factor of $\sqrt[n]{m}$ which goes to $1$ as $n\to\infty$.

Answer 2

Let $M=\max (x_1,...,x_k)=x_i.$ Then $$M=x_i=(x_i^n)^{1/n}\leq (\sum_{j=1}^kx_j^n)^{1/n}=a_n\leq (\sum_{j=1}^kx_i^n)^{1/n}=(x_i^n\cdot k)^{1/n}=x_i\cdot k^{1/n}=Mk^{1/n}. $$