# Comparing growth of sequences in weighted spaces

I would like to ask a follow-up question on a previous question 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 $$x\in A$$

$$\limsup_{\varepsilon \downarrow 0} \frac{F_{\varepsilon}(n)}{F_{\varepsilon}(x)} \le 1?.$$

$$\newcommand\ep\varepsilon\newcommand\de\delta$$ 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 $$N>M_\de$$, $$\begin{equation*} \sum_1^N n^2 y_n=\sum_{n\le M_\de} n^2 y_n+\sum_{M_\de if we also have $$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.