For positive $a_1,a_2,a_3......,a_p $ Find $\lim_{n\rightarrow \infty} {\sqrt[n] \frac{a_1^n +a_2^n +........+a_p^n}{p}}$ 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 .....
 A: 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 $$\lim_{n \to \infty}\sqrt[n]{\frac{a_1^n+a_2^n+\cdots+a_p^n}{p}}=\lim_{n \to \infty}\frac{\sqrt[n]{a_1^n+a_2^n+\cdots+a_p^n}}{\sqrt[n]{p}}=M.$$
A: 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.
A: 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!
