Asymptotic behavior of the moments of non-negative sequences

Question

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 $\sum_{k=1}^n k^p u_k$.

Answer 1

$\newcommand\ep\epsilon$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 $n$.