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 .....

Answer 1

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

Answer 2

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.

Answer 3

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!