Existence of a sequence with bounded averages I wonder if there exist a sequence $(\beta_{i})$ in positive real numbers such that
$$\sup_{p} (\frac{1}{p}\sum_{i=1}^{p}\beta_{i})<\infty$$, $$\sup_{p}(\frac{1}{p}\sum_{i=1}^p \beta_i^2)<\infty$$ and $$\sum_{i=1}^{\infty}\frac{\beta_{i}^2}{i^2}=\infty$$.
So far I could not find the existence or could not prove the nonexistence of such sequence. Any help is appreciated.
 A: Let $(\beta_n)_{n\geq 1}$ be a sequence of real numbers. Write $T_n = \frac{1}{n}\sum_{i=1}^{n} \beta_i^2$ and assume that $(T_n)_{n\geq 1}$ is bounded. Then by the identity
$$ \sum_{i=1}^{n} \frac{\beta_i^2}{i^2}
= \frac{T_n}{n} + \sum_{k=1}^{n-1} \frac{2k+1}{k(k+1)^2} T_k, $$
we get
$$\sum_{i=1}^{\infty} \frac{\beta_i^2}{i^2}
= \sum_{k=1}^{\infty} \frac{2k+1}{k(k+1)^2} T_k
< \infty. $$
A: @SangchulLee's solution is sharp, I am impressed. It leaves room for a more naive solution.
We already know from the mentioned solution that the first assumption is not needed. Thus each $\ \beta_k^2\ $ can be replaced by a non-negative $\ a_k.\ $ Thus let's formulate and prove
THEOREM For each sequence of non-negative reals $\ a_n,\ $
and for arbitrary $\ M\in\mathbb R\ $ such that
$$ \forall_{n\in\mathbb N}\quad \frac 1n\cdot\sum_{k=1}^n a_k
    \ \le\ M $$
also
$$ \sum_{k\in\mathbb N}\frac{a_k}{k^2}\ <\ \infty $$
holds.
PROOF
$$ \forall_{n\in\mathbb N}\quad
   \sum_{k=n}^{2\cdot n-1} \frac{a_k}{k^2}\ \le
\ \frac 1n\cdot\big(\frac 1n\cdot\sum_{k=n}^{2\cdot n-1} a_k\big)
    \ \le\ \frac {2\cdot M}n $$
i.e.
$$ \forall_{n\in\mathbb N}\quad
   \sum_{k=n}^{2\cdot n-1} \frac{a_k}{k^2}\ \le
\ \frac {2\cdot M}n $$
Thus
$$ \sum_{k\in\mathbb N}\ \frac{a_k}{k^2}\,\ =\,
  \ \sum_{m=0}^\infty\ \sum_{k=2^m}^{2^{m+1}-1}\ \frac{a_k}{k^2}
\,\ \le\,\ 2\cdot M\cdot\sum_{m=0}^\infty 2^{-m}
\,\ =\,\ 4\cdot M\,\ <\,\ \infty $$
 
End Of PROOF

P.S. In a comment below, OP @Hubeyb has asked about the inverse of the above theorem; and just below it I've written "no" but now in more words:

Let $\ \forall_{k\in\mathbb N}\ a_k:=\sqrt k.\ $ Then the inverse fails, namely:
$$ \lim_{n=\infty} \frac 1n\cdot\sum_{k=1}^n \sqrt k\ =\ \infty $$
and
$$ \sum_{k=1}^\infty\ \frac{\sqrt k}{k^2}\ <\ \infty $$
Thus the above theorem works in that given direction only, and not in the opposite direction.
