For positive $a_1,a_2,a_3......,a_p $ Find $\lim_{n\rightarrow \infty} {\sqrt[n] \frac{a_1^n +a_2^n +........+a_p^n}{p}}$

Question

For positive $a_1,a_2,a_3......,a_p $

Find $\lim_{n\rightarrow \infty} {\sqrt[n] \frac{a_1^n +a_2^n +........+a_p^n}{p}}$

My attempts :${\sqrt[n] \frac{a_1^n +a_2^n +........+a_p^n}{p}}$= ${ \sqrt[n] {a_1^n +a_2^n +........+a_p^n}}. (\frac{1}{\sqrt[n]{p}} )$

$\sqrt[n]{a_1^n+\cdots+a_p^n}\leq \sqrt[n]{a_p^n+\cdots+a_{p}^n}=\sqrt[n]{na_p^n}=\sqrt[n]{n} \cdot a_p$

so ${\sqrt[n] \frac{a_1^n +a_2^n +........+a_p^n}{p}}$= ${ \sqrt[n] {a_1^n +a_2^n +........+a_p^n}}. (\frac{1}{\sqrt[n]{p}} )\le \frac {\sqrt[n]{n} \cdot a_p} (\frac{1}{\sqrt[n]{p}} $

after that I'm not able to proceed further .....

Pliz help me

thanks in advance .....

What about using the Binomial Theorem? $$(a_1+a_2+\cdots + a_p)^n = (a_1^n+\cdots + a_p^n + \text{ other stuff})$$ Maybe you can simplify it from there? — SlipEternal, Jul 10 '18 at 12:36
Let $a=\max(a_1,a_2,...,a_p)$, then write $\sqrt[n]{\frac{a_1^n+...+a_p^n}{p}}=a\sqrt[n]{\frac{(a_1/a)^n+...+(a_p/a)^n}{p}}$ Take into account that when $a_i=a$, then $a_i/a=1$, otherwise $0<a_i/a<1$. — , Jul 10 '18 at 12:37
The limit is $\max_ia_i$ this is a standard argument of $p$-means (see f.i. the book of Hardy Littlewood and Polya) — Surb, Jul 10 '18 at 12:40
Possible duplicate of Why is the $0$th power mean defined to be the geometric mean? — Isaac Browne, Jul 10 '18 at 21:14

score 7 · Accepted Answer · edited Jun 12 '20 at 10:38

7

Solution

Let $$\max(a_1,a_2,\cdots,a_p)=M.$$

Then $$ M=\sqrt[n]{M^n}\leq \sqrt[n]{a_1^n+a_2^n+\cdots+a_p^n}\leq \sqrt[n]{p\cdot M^n}=M\cdot\sqrt[n]{p}.$$

Notice that $\sqrt[n]{p} \to 1$ as $n \to \infty$. Hence, by the squeeze theorem, we may obtain $$\lim_{n \to \infty} \sqrt[n]{a_1^n+a_2^n+\cdots+a_p^n}=M.$$

It follows that $$\lim_{n \to \infty}\sqrt[n]{\frac{a_1^n+a_2^n+\cdots+a_p^n}{p}}=\lim_{n \to \infty}\frac{\sqrt[n]{a_1^n+a_2^n+\cdots+a_p^n}}{\sqrt[n]{p}}=M.$$

edited Jun 12 '20 at 10:38

Community

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answered Jul 10 '18 at 12:55

WuKong

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thanks u @mengdie – jasmine Jul 10 '18 at 12:57
This Q reminds me of the exercise: If $f:[0,1]\to \Bbb R$ is continuous then $\lim_{p\to \infty}(;\int_0^1|f(x)|^pdx;)^{1/p}=$ $\max {|f(x)|: x\in [0,1]}.$ – DanielWainfleet Jul 10 '18 at 18:09

score 2 · Answer 2 · answered Jul 10 '18 at 12:42

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since $p>1$ is fixed, $\sqrt[n]{p}\to 1$, so we can ignore the factor of $1/\sqrt[n]{p}$. Set $M = \max(a_1,\dots,a_p)$, and without loss the $a_i$ are non-increasing. Then $$ \sqrt[n]{a_1^n +a_2^n +........+a_p^n} = M \sqrt[n]{1 +b_2^n +........+b_p^n} $$ with $b_i^n = a_i^n/a_1^n \le 1$, so $$ M ≤ \sqrt[n]{a_1^n +a_2^n +........+a_p^n} ≤ M\sqrt[n]{p} \to M$$ so the limit is M.

answered Jul 10 '18 at 12:42

Calvin Khor

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thanks u calvin khor – jasmine Jul 10 '18 at 12:57

AD - Stop Putin - · Answer 3 · 2018-07-10T12:57:52.060

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This part is wrong (unless you have a growing sequence): $\sqrt[n]{a_1^n+\cdots+a_p^n}\leq \sqrt[n]{a_p^n+\cdots+a_{p}^n}=\sqrt[n]{na_p^n}=\sqrt[n]{n} \cdot a_p$

However, the left-hand side is bounded by

$$\sqrt[n]{p \max a_k^n}= \sqrt[n]{p} \cdot \max a_k $$ (Prove it!)

This is actually the limit, and in order to prove that all that is needed is to find an estimate of $$a_1^n+\cdots+a_p^n$$ from below. You can do that, I promis!

edited Jul 10 '18 at 12:57

answered Jul 10 '18 at 12:53

AD - Stop Putin -

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thanks u @AD.... – jasmine Jul 10 '18 at 12:57
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No problem!! – AD - Stop Putin - Jul 10 '18 at 12:59

For positive $a_1,a_2,a_3......,a_p $ Find $\lim_{n\rightarrow \infty} {\sqrt[n] \frac{a_1^n +a_2^n +........+a_p^n}{p}}$

3 Answers3

Solution