Tauberian lower bound for a series Let $(a_n)_{n\in\mathbb{N}}$ be a sequence of positive number such that $\sum_n a_n < +\infty$ (i.e. $a_n \in \ell^1$) but $\sum_n r^n a_n = +\infty$ for every $r > 1$.
Given $\sigma \in (0,1)$, I would like to prove a lower bound on the function which maps $t \geq 1$ to
$$
H_\sigma(t) := \sum_n \frac{4^{n \sigma}}{4^n+t^2} a_n.
$$
Concerning an upper bound, one easily proves that $H_\sigma(t) \leq t^{-2(1-\sigma)}$.
Concerning the lower bound, one easily gets similar behaviors for particular sequences. For example, when $a_n = 1/n^2$ (which satisfies the assumptions), taking heuristically only the term for $n = \lfloor \ln t \rfloor$ one obtains $H_\sigma(t) \geq t^{-2(1-\sigma)} / \ln^2(t)$.
I suspect some kind of Tauberian argument could yield a lower bound valid for all sequences $a_n$ as in the assumptions, but I am not familiar with such techniques. I would be happy with a result of the form:
$$\forall \sigma' < \sigma, \exists c' >0, \quad H_\sigma(t) \geq c' t^{-2(1-\sigma')}.$$
Could you suggest a method or references on related arguments?
 A: $\newcommand{\si}{\sigma}\newcommand{\ep}{\varepsilon}$The answer is no. If the sequence $(a_n)$ is lacunary enough, then $H_\si(t)$ may behave for some large $t$ roughly as just one of the summands $\frac{4^{n \si}}{4^n+t^2} a_n$, so that for such $t$ we have $H_\si(t)\ll t^{-2+\ep}$ for every real $\ep>0$, and hence, for any given $\si'>0$, the inequality $H_\si(t)\gg t^{-2(1-\si')}$ will not hold for all large enough $t$. We write $A\ll B$ and $B\gg A$ if $A=O(B)$.
To simplify notations, let
\begin{equation*}
    u:=t^2,\quad s:=\si,\quad s':=\si', 
\end{equation*}
so that
\begin{equation*}
    H_\si(t)=h_s(u):=\sum_n \frac{4^{sn}}{4^n+u}\, a_n. \tag{1}\label{1}
\end{equation*}
Take a natural number $b\in\{3,4,\dots\}$, which should be thought of as fixed but large enough (depending only on $s$), as large as needed wherever needed. Let
\begin{equation*}
    a_n:=\frac1{b^j}\text{ if $n=b^j$ for some natural $j$, with $a_n=0$ otherwise.} 
\end{equation*}
Then $\sum_n a_n<\infty$ and $\sum_n r^n a_n=\infty$ for every $r>1$. Moreover,
\begin{equation*}
h_s(u)\le g_s(u):=\sum_j c_j(u), \tag{2}\label{2}
\end{equation*}
where
\begin{equation*}
    c_j(u):=\frac{4^{sb^j}}{4^{b^j}+u}. 
\end{equation*}
Consider
\begin{equation*}
    r_j(u):=\frac{c_{j+1}(u)}{c_j(u)}=4^{s(b-1)b^j}\frac{4^{b^j}+u}{4^{b\,b^j}+u}. 
\end{equation*}
Finally, choose
\begin{equation*}
    u=u_k:=4^{b^k[(1-s)b+s]/2},
\end{equation*}
where $k$ is a natural number going to $\infty$, so that $u_k\to\infty$.
Note that $1\le[(1-s)b+s]/2\le b$ for all large enough $b$ (depending only on $s$) and hence
\begin{equation*}
    4^{b^k}\le u\le4^{b\,b^k}. \tag{3}\label{3}
\end{equation*}
So, for natural $j\ge k$ we have
\begin{equation*}
    r_j(u)\le4^{s(b-1)b^j}\frac{4^{b^j}+u}{4^{b\,b^j}}
    \le4^{-(1-s)(b-1)b^k}+4^{-b^k[(1-s)b+s]/2}\le\frac12
\end{equation*}
for all large enough $b$.
So,
\begin{equation*}
\sum_{j\ge k} c_j(u)\le2c_k(u). \tag{4}\label{4}
\end{equation*}
On the other hand, for natural $j\le k-1$ we have
\begin{equation*}
    r_j(u)\ge4^{s(b-1)b^j}\frac{u}{4^{b^k}+u}\ge4^{s(b-1)b^j}\frac12\ge2
\end{equation*}
for all large enough $b$; the penultimate inequality follows by \eqref{3}.
So,
\begin{equation*}
\sum_{j\le k-1} c_j(u)\le2c_k(u). \tag{5}\label{5}
\end{equation*}
By \eqref{2}, \eqref{4}, and \eqref{5},
\begin{equation*}
h_s(u_k)\le 4c_k(u_k)\le4\frac{4^{sb^k}}{u_k}=4u_k^{-p},
\end{equation*}
where
\begin{equation}
    p:=\frac{(1-s)b-s}{(1-s)b+s},
\end{equation}
which can be made however close to $1$ by letting $b$ be large enough.
Thus, for any real $s'>0$, the statement $h_s(u)\gg u^{-(1-s')}$ will fail to hold, if $b$ is large enough. Equivalently, for any real $s'>0$ and any real $c'>0$, it is not true that $H_s(t)\ge c' t^{-2(1-s')}$ for all large enough $t>0$. $\quad\Box$
