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?

Solution 1:

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?