Find a sequence limit 
Assume $a_n>0$ and $\lim\limits_{n \to
\infty}\dfrac{a_1+a_2+\cdots+a_n}{n}=a<+\infty$. Find $$\lim\limits_{n
 \to \infty}\dfrac{a_1^p+a_2^p+\cdots+a_n^p}{n^p}$$ where $p>1$.

Consider applying Stolz's theorem, we obtain
$$\lim\limits_{n \to \infty}\dfrac{a_1^p+a_2^p+\cdots+a_n^p}{n^p}=\lim_{n \to \infty}\frac{a_{n+1}^p}{n^p\left[\left(1+\frac{1}{n}\right)^p-1\right]}=\lim_{n \to \infty}\frac{a_{n+1}^p}{pn^{p-1}}.$$
This will help?
 A: Someone formulates a solution under the assumption $p>1$.
Let $q:=p-1$. Then
$$\lim_{n \to \infty} \frac{a_n}{n}=\lim_{n \to \infty}\left(\frac{a_1+\cdots+a_{n}}{n}-\frac{n-1}{n}\cdot\frac{a_1+\cdots+a_{n-1}}{n-1}\right)=a-1\cdot a=0. $$
Define $b_n:=a_n+1$. It hold that
$$\lim_{n \to \infty}(b_1+b_2+\cdots+b_n)=+\infty, $$ and$$\lim_{n \to
   \infty} \frac{b_n^q}{n^q}=\lim_{n \to
   \infty}\left(\frac{b_n}{n}\right)^q=\left[\lim_{n \to
   \infty}\left(\frac{a_n}{n}+\frac{1}{n}\right)\right]^q=0. $$
As per the weighted Cauchy's limit theorem, we have
$$\lim_{n \to \infty}\dfrac{b_1\cdot\frac{b_1^q}{1^q}+b_2\cdot\frac{b_2^q}{2^q}+\cdots+b_n\cdot\frac{b_n^q}{n^q}}{b_1+b_2+\cdots+b_n}=0. $$
Moreover, since $\left\{\dfrac{b_n}{n}\right\}$ is convergent, hence bounded, there exists some $C>0$ such that $b_1+b_2+\cdots+b_n<Cn$. Therefore
$$0\le\frac{a_1^p+a_2^p+\cdots+a_n^p}{Cn^p}\le\frac{b_1^p+b_2^p+\cdots+b_n^p}{Cn^p}\le\dfrac{b_1\cdot\frac{b_1^q}{1^q}+b_2\cdot\frac{b_2^q}{2^q}+\cdots+b_n\cdot\frac{b_n^q}{n^q}}{b_1+b_2+\cdots+b_n},$$
which implies
$$\lim_{n \to \infty}\frac{a_1^p+a_2^p+\cdots+a_n^p}{n^p}=0,$$
by the squeezing theorem.
This is correct?
A: @mengdie1982 Your proof seems valid in my view. I thought  of another proof which uses some of your notations (I didn't use "weighted Cauchy's limit theorem" which I'm not familiar with).
using your notation,  denote $q=p-1>0$ and using the fact that you proved:    $\lim\limits_{n\rightarrow\infty} \frac {a_n} n=0$
notice that
$\frac {a^p_1+a^p_2+++a^p_n}{n^p}=\left(\frac {a_1} n \right) ^p+\left(\frac {a_2} n\right) ^p+++\left(\frac {a_n} n\right) ^p=\frac {a_1} n \left(\frac {a_1} n \right) ^q+\frac {a_2} n \left(\frac {a_2} n \right) ^q+++ \frac {a_n} n \left(\frac {a_n} n \right) ^q < \left(\frac {a_1} n + \frac {a_2} n +++ \frac {a_n} n \right)*\left(\frac {\max\limits_{i\leq n} (a_i)} n\right) ^q $
Notice that $\lim\limits_{n\rightarrow\infty} \frac {a_1} n + \frac {a_2} n +++ \frac {a_n} n=a$ and so in order to prove that the last expression has a limit 0 it is sufficiant to prove that $\lim\limits_{n\rightarrow\infty} \frac {\max\limits_{i\leq n} (a_i)}n=0$
So, denote by $\max_n$ as the index of the element $a_i$ such that $a_{\max_n}=\max\limits_{i\leq n} (a_i)$. Notice that $\max_n$ is a non decreasing fuction of n.
Notice that $\frac{\max\limits_{i\leq n} (a_i)}n=\frac{a_{\max_n}}n=\frac{a_{\max_n}}{\max_n}*\frac {\max_n}n$
Now, since $\max_n$ is a non decreasing fuction of n there are 2 posibilites: either $\lim\limits_{n\rightarrow\infty} \max_n=\infty$ and in this case $\lim\limits_{n\rightarrow\infty} \frac{a_{\max_n}}{\max_n}=0$ because we noticed that $\lim\limits_{n\rightarrow\infty} \frac {a_n} n=0$. The other posibility is $\max_n=CONST$ for all n>N for some N and in this case $\lim\limits_{n\rightarrow\infty} \frac {\max_n}n=0$. So we conclude that $\lim\limits_{n\rightarrow\infty} \frac {\max\limits_{i\leq n} (a_i)}n=0$
