Let $(a_n)$ a sequence of positive numbers such that $\lim\dfrac{a_1+…+a_n}{n}<\infty$, $\lim \dfrac{a_n}{n}=0$. Prove that $\lim\dfrac{a_1^2+…+a_n^2}{n^2}= 0$.
My attempt:
Call $\lim \dfrac{a_1+…+a_n}{n} = L$.
Let $\epsilon>0$. I already proved that $L = \lim a_n$. Then, since $x^2$ is continuous, $L^2 = \lim a_n^2$. Then there is $n_0$ such that $n>n_0$ implies
$|a_n^2 – L^2|<\frac{\epsilon}{2}$ and $a_n^2$ is convergent then limited by $M$.
If I can say that $\lim a_n^2 = 0$, then (I think) I'm done, since:
$0\leq\displaystyle\left|\sum_{k=1}^{n}\dfrac{a_k^2}{n^2}\right|=\sum_{k=1}^{n}\dfrac{a_k^2}{n^2}=\sum_{k=1}^{n_0}\dfrac{a_k^2}{n^2}+\sum_{k=n_0+1}^{n}\dfrac{a_k^2}{n^2}\leq \dfrac{1}{n^2}\sum_{k=1}^{n_0}M + \dfrac{1}{n^2}\sum_{k=n_0+1}^{n}a_k^2=\dfrac{Mn_0}{n^2} + \dfrac{1}{n^2}\sum_{k=n_0+1}^{n}a_k^2<\dfrac{Mn_0}{n^2} + \dfrac{1}{n^2}\sum_{k=n_0+1}^{n}\dfrac{\epsilon}{2}=\dfrac{Mn_0}{n^2} + \dfrac{\epsilon(n-n_0)}{2n^2}<\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon$
Where I choosed $n_1>n_0$ such that $n>n_1$ implies $\dfrac{Mn_0}{n^2}<\dfrac{\epsilon}{2}$
But how can I prove $\lim a_n^2=0$, if it is true? I think $\lim \dfrac{a_n}{n}=0$ may imply that, but I also think this hypothesis is very odd, since $\lim \dfrac{a_1+…+a_n}{n} = \lim an = L$, so $\lim \dfrac{a_n}{n} = 0$ should be an obvious consequence…
Best Answer
You started out correctly.
To finish, since $a_k/k \to 0$ note that there exists $n_0$ such that $0 < a_k < k \epsilon \leqslant n \epsilon$ if $n \geqslant k > n_0$, and
$$0 \leqslant \frac{1}{n^2}\sum_{k=1}^{n}a_k^2= \frac{1}{n^2}\sum_{k=1}^{n_0}a_k^2 + \frac{1}{n^2}\sum_{k=n_0+1}^{n}a_k^2 \leqslant \frac{1}{n^2}\sum_{k=1}^{n_0}a_k^2 + \frac{n\epsilon}{n^2}\sum_{k=n_0+1}^{n}a_k $$
(The last inequality follows because $0 < a_k < n\epsilon$ implies $ n\epsilon a_k < a_k^2$ for $k > n_0$.)
Thus, for any $\epsilon > 0$ we have
$$0 \leqslant \liminf \frac{1}{n^2}\sum_{k=1}^{n}a_k^2 \leqslant \limsup \frac{1}{n^2}\sum_{k=1}^{n}a_k^2 < \epsilon \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^na_k = \epsilon L,$$
which implies
$$\lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^na_k^2 = 0$$