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Can someone show me how to proof this ? $$\sqrt[p]{\sum_i {a_i}^p}\to \max a_i \quad\text{ if }p\to\infty.$$

Muaa2404
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  • guess there is a $1/n$ missing for this to be true... – GaC Oct 30 '16 at 15:53
  • @SpettroDiA He wrote $\sqrt[p]{\text{stuff}}$. Perhaps, what's missing is the absolute value of the $a_i$. And a $\sup$ instead of $\max$, if we stand by the tag "lp-spaces" –  Oct 30 '16 at 15:54
  • There is no need for 1/n. This limit is also known as Chebyshev distance. Thank for help. – Muaa2404 Oct 30 '16 at 15:58
  • This is false for general sequences. Simply take $a_n = 1$ for all $n$ to see this. If we assume $(a_n) \in l^p$ for some $p>0,$ it's true. – zhw. Oct 30 '16 at 16:39

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Assuming $\forall i, a_i \le 0$

We have :$$\sum_i a_i^p \sim k \max_i a_i^p$$ where $k$ is the number of times the number $\max_i a_i$ appears in the sum.

Thus $$\sqrt[p]{\sum_i a_i^p}\sim \sqrt[p]k\max_i{a_i} \sim \max_i{a_i}$$

(since $k\ge1$)

Astyx
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  • It is a finite sum of geometric sequences, not a series. The preponderant term is the geometric sequence with the greatest value (all others are negligible) – Astyx Oct 30 '16 at 15:58
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    Well, the tag was "lp-spaces". However, I guess he still uses $\max$ instead of $\sup$. –  Oct 30 '16 at 16:00