We consider a sequence $u = (u\_k)\_{k\geq 1}$ such that $u\_k \geq 0$ for any $k \geq 1$. We assume that there exists a critical $p\_c \in \mathbb{R}$ such that, for any $q<p\_c <p$,
$$\sum\_{k=1}^\infty k^q u\_k < \infty \quad \text{and} \quad \sum\_{k=1}^\infty k^p u\_k = \infty.$$
I am interested in the asymptotic behavior of $\sum\_{k=1}^n k^p u\_k$ when $n\rightarrow \infty$ for $p > p\_c$.


More precisely, I expect this quantity to behaves roughly as $n^{p-p\_c}$ in the following sense.


**Conjecture:** For any $p > p\_c$ and $0 < \epsilon < p - p\_c$, there exists $0 < m , M < \infty$ such that, for any $n \geq 1$,
$$m n^{p - p\_c - \epsilon} \leq \sum\_{k=1}^n k^p u\_k \leq M n^{p - p\_c + \epsilon}.$$


The upper bound of the conjecture is easy to obtain. Indeed, we readily have that
$$\sum\_{k=1}^n k^p u\_k \leq \left( \sum\_{k=1}^n k^{p\_c - \epsilon} u\_k \right) n^{p-p\_c + \epsilon} \leq \left( \sum\_{k=1}^\infty k^{p\_c - \epsilon} u\_k \right) n^{p-p\_c + \epsilon},$$
hence the constant $M = \sum\_{k=1}^\infty k^{p\_c - \epsilon} u\_k$ works.


My question is: Is the conjecture true in the sense that the constants $m$ always exist to lower-bound $\sum\_{k=1}^n k^p u\_k$.



Asymptotic behavior of the moments of non-negative sequences
$\newcommand\ep\epsilon$The answer is no.


Indeed, let e.g.
$u\_k:=1$ if $k=k\_j:=2^{5^j}$ for some natural $j$, and $u\_k:=0$ otherwise. Then $p\_c=0$.


Take now any real $\ep>0$ and then take $p=2\ep$, so that condition $0< \epsilon <p-p\_c$ holds. Then for all large enough $j$ and $n=k\_{j+1}-1$ we have
$$n^{p-p\_c-\ep}=(k\_{j+1}-1)^\ep\ge2^{\ep 5^{j+1}/2}$$
and
$$\sum\_{k=1}^n k^pu\_k=\sum\_{i=1}^j k\_i^{2\ep}
\le2k\_j^{2\ep}=2\times2^{2\ep 5^j}
=o(2^{\ep 5^{j+1}/2})=o(n^{p-p\_c-\ep})$$
as $j\to\infty$.


So, there is no real $m>0$ such that $mn^{p-p\_c-\ep}\le\sum\_{k=1}^n k^pu\_k$ for all $n$.


