Limit $\lim_{n \rightarrow \infty} (A_1^n + ... A_k^n)^{1/n}= \max\{ A_1, ..., A_k\}$

Question

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?

https://math.stackexchange.com/questions/326172/the-l-infty-norm-is-equal-to-the-limit-of-the-lp-norms — Guy Fsone, Jan 28 '18 at 11:26

score 2 · Accepted Answer · answered Oct 29 '14 at 17:21

2

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

answered Oct 29 '14 at 17:21

Yiorgos S. Smyrlis

86,581

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

score 0 · Answer 2 · answered Oct 29 '14 at 17:12

0

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 $b_n^{1/n}$.

answered Oct 29 '14 at 17:12

mookid

28,795

Limit $\lim_{n \rightarrow \infty} (A_1^n + ... A_k^n)^{1/n}= \max\{ A_1, ..., A_k\}$

2 Answers2

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