I would like to ask a follow-up question on a [previous question](https://mathoverflow.net/questions/386802/is-the-harmonic-series-worse-than-any-summable-series) of mine here whose proof does not seem to carry over to this case in an obvious way:


We define the function $$F\_{\varepsilon}(x) = \sum\_{i=1}^{\infty} 2^{-\varepsilon \vert x\_i \vert} \text{ for }\varepsilon>0.$$


We let $A$ be the set of positive sequences $x=(x\_n)$ such that $\sum\_n \frac{x\_n}{n^2}<\infty$. Clearly, the sequence $x=(n)$ is not in $A$.


Since we expect that anything in $A$ cannot grow as fast as $x=(n)$, I ask: Is it true that for any sequence $x\in A$


$$\limsup\_{\varepsilon \downarrow 0} \frac{F\_{\varepsilon}(n)}{F\_{\varepsilon}(x)} \le 1?.$$



Comparing growth of sequences in weighted spaces
$\newcommand\ep\varepsilon\newcommand\de\delta$
Let us show more: for all $x\in A$,
\begin{equation\*}
 \frac{F\_{\ep}((n))}{F\_{\ep}(x)}\to0\tag{$\*$}
\end{equation\*}
(as $\ep\downarrow0$).


Indeed, take any $x\in A$ and let
\begin{equation\*}
 y\_n:=x\_n/n^2,
\end{equation\*}
so that $\sum\_n y\_n<\infty$ and
\begin{equation\*}
 F\_{\ep}(x)=\sum\_{n=1}^\infty 2^{-\ep n^2 y\_n}. 
\end{equation\*}
By Jensen's inequality for the convex function $u\mapsto2^{-u}$, for any natural $N$
\begin{equation\*}
 F\_{\ep}(x)\ge\sum\_{n=1}^N 2^{-\ep n^2 y\_n}\ge N2^{-\ep \sum\_1^N n^2 y\_n/N}. \tag{1}
\end{equation\*}
Take now any real $\de>0$. Then, by the condition $\sum\_n y\_n<\infty$, there is a natural $M\_\de$ such that $\sum\_{n>M\_\de} y\_n<\de/2$. So, for $N>M\_\de$,
\begin{equation\*}
 \sum\_1^N n^2 y\_n=\sum\_{n\le M\_\de} n^2 y\_n+\sum\_{M\_\de<n\le N} n^2 y\_n
 \le\sum\_{n\le M\_\de} n^2 y\_n+N^2\de/2<N^2\de
\end{equation\*}
if we also have $N^2>2\sum\_{n\le M\_\de} n^2 y\_n/\de$, and then, by (1),  

\begin{equation\*}
 F\_{\ep}(x)\ge N2^{-\ep N\de}. 
\end{equation\*}
Choosing now $N\sim\dfrac1{\ep\de}$ with $\ep\downarrow0$, we have
\begin{equation\*}
 F\_{\ep}(x)\ge\dfrac1{3\ep\de}, 
\end{equation\*}
for each real $\de>0$ and all small enough $\ep>0$.


On the other hand,
\begin{equation\*}
 F\_{\ep}((n))=\sum\_{n=1}^\infty 2^{-\ep n}=\frac{2^{-\ep}}{1-2^{-\ep}}\sim\frac1{\ep\ln2}. 
\end{equation\*}


Now ($\*$) follows.


