so I have a series of real numbers $(q_n)_{n\in\mathbb N}$ ($q_i \in [0,1]$) depending on some $\alpha \in (0,1/2)$ and what I know is the following, there exists constants $c_1, c_2$ so that:
$q_i \sim c_1 i^{\alpha -1}$ for $i\rightarrow \infty$
and
$\sum_{i=0}^k q_i \sim c_2 k^{\alpha}$ for $k\rightarrow \infty$
holds.
I would be interested if there is something like a Tauberian Theorem that could give one something like
$\sum_{i\geq 0} q_i q_{i+j} \sim c_3j^{2 \alpha -1} $ for $j\rightarrow \infty$
for some constant $c_3$ or something like that? It would be even better to get asymptotic results for $\sum_{i=0}^N q_i q_{i+j}$? Maybe there's also something?
Would be really glad if someone has a hint or an idea!
All the best!
If $q_k \sim C k^{-r}$ with $r\in (1/2,1),C\ne 0$ then $$\sum_{k\ge 1}q_k q_{k+j} = O(\sum_{k=1}^{\log j} k^{-r} (k+j)^{-r})+ \sum_{k>\log j} C^2 k^{-r} (k+j)^{-r}(1+o(1))$$ $$= O( (\log j)^{1-r}j^{-r} )+(C^2+o(1))\sum_{k>\log j} k^{-r} (k+j)^{-r}$$ Then $\sum_{k>\log j} k^{-r} (k+j)^{-r}>\sum_{k>\log j}^{j} k^{-r} (2j)^{-r}\sim (2j)^{-r}\frac{ j^{1-r}}{1-r}$ so
$$\sum_{k\ge 1}q_k q_{k+j} = (C^2+o(1))\sum_{k>\log j} k^{-r} (k+j)^{-r} \sim C^2\sum_{k>1} k^{-r} (k+j)^{-r}$$ I don't know how to find the asymptotic of the latter series, but it is close to $j^{1-2r}$.