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A root limit

Let a_i \; , \; i =1, 2, \dots, k be positive real numbers such that a_1\geq a_2\geq \cdots \geq a_k. Prove that

    \[\lim_{n \rightarrow +\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_k^n} = \max\left \{ a_1, a_2 , \dots, a_k \right \}\]

Solution

Without loss of generation , let a_1 =\max\{ a_1, a_2 , \dots , a_k \}. Then,

    \begin{align*} a_1 &=(a_1^n)^{1/n}\\ &\leq (a_1^n+\cdots+a_k^n)^{1/n}\\ &=\sqrt[n]{a_1^n \left (1+\left (\frac{a_2}{a_1} \right )^n + ... + \left (\frac{a_k}{a_1} \right )^n \right )} \\ &\leq \sqrt[n]{{a_1}^n \cdot k} \\ &= a_1\sqrt[n]{k} \\ &=a_1 k^{1/n} \end{align*}

since a_i \leq  a_1 forall i=1,2, \dots,k. Thus, by the squeeze theorem it follows that

    \[\lim_{n \rightarrow +\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_k^n} =a_1 =  \max\left \{ a_1, a_2 , \dots, a_k \right \}\]

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