# Limit $\lim_{n \rightarrow \infty} (A_1^n + … A_k^n)^{1/n}= \max\{ A_1, …, A_k\}$ [duplicate]

I have the following question. I was asked to compute the following limit: Let $A_1 ... A_k$ be positive numbers, does exist:

$$\lim_{n \rightarrow \infty} (A_1^n + ... A_k^n)^{1/n}$$ My work: W.L.O.G let $A_1= \max{ A_1, ..., A_k}$, so I have $$A_1^n \leq A_1^n + ... A_k^n \leq kA_1^n$$ so that

$$A_1 = \lim_{n \rightarrow \infty} (A_1^n)^{1/n} \leq \lim_{n \rightarrow \infty}(A_1^n + ... A_k^n)^{1/n} \leq \lim_{n \rightarrow \infty} (kA_1^n)^{1/n} = kA_1$$

Can I do something else to sandwich the limit?

## marked as duplicate by Guy Fsone, Fabio Lucchini, muaddib, Parcly Taxel, Hans EnglerJan 28 '18 at 13:59

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## 2 Answers

You have made a little mistake. Correction: $$A_1 = \lim_{n \rightarrow \infty} (A_1^n)^{1/n} \leq \lim_{n \rightarrow \infty}(A_1^n + ... A_k^n)^{1/n} \leq \lim_{n \rightarrow \infty} (kA_1^n)^{1/n} = \lim_{n \rightarrow \infty} A_1{\color{blue}{k^{1/n}}}=A_1$$

• Gee thanks for pointing that out. – David Cardozo Oct 29 '14 at 18:05

Hint: $$(ab_n)^{1/n} = \exp \left(\frac 1n \log b_n + \frac 1n \log a \right)$$ has the same limit when $n\to\infty$ (if any) as $b_n^{1/n}$.