It is well-known that the harmonic series is not summable. In some sense this means that it takes a lot of rather large values.


We define the operator $F\_{\varepsilon}: \ell^{\infty}(\mathbb N) \rightarrow [0,\infty]$ by $$F\_{\varepsilon}(x) = \sum\_{i=1}^{\infty} 2^{-\varepsilon \vert x\_i \vert^{-1}} \text{ for }\varepsilon>0.$$


Now consider a positive summable sequence $x$ and the harmonic sequence $(1/n)\_n$. Intuitively, the slow decay of the harmonic series should imply that it converges slower than anything summable (for most of it).


Therefore, I ask: Is it true that for any positive summable sequence $x$


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


Please just let me know if you have any questions.



Is the harmonic series worse than any summable series?
$\newcommand\ep\varepsilon\newcommand\de\delta$
Let us show more:
\begin{equation\*}
 \frac{F\_{\ep}(x)}{F\_{\ep}((1/n))}\to0\tag{$\*$}
\end{equation\*}
(as $\ep\downarrow0$).
Indeed,
\begin{equation\*}
 F\_{\ep}((1/n))=\sum\_{i=1}^\infty 2^{-\ep i}=\frac{2^{-\ep}}{1-2^{-\ep}}\asymp\frac1\ep. \tag{0}
\end{equation\*}


On the other hand, take any positive $x\_n$'s such that $\sum\_1^\infty x\_n<\infty$. For each natural $k$, let $J\_k$ denote the set of all natural $n$ such that $\frac1k\le x\_n<\frac1{k-1}$, where $\frac1{k-1}:=\infty$ for $k=1$:
\begin{equation\*}
 n\in J\_k\iff\frac1k\le x\_n<\frac1{k-1}. 
\end{equation\*}
Then the $J\_k$'s partition the set of all natural numbers. Moreover, the condition $\sum\_1^\infty x\_n<\infty$ implies
\begin{equation\*}
 \sum\_{k=1}^\infty|J\_k|/k<\infty, \tag{1}
\end{equation\*}
where $|J\_k|$ is the cardinality of $J\_k$.
In particular, it follows that $|J\_k|<\infty$ for all $k$.
Further, $2^{-\ep/x\_n}<2^\ep\times2^{-\ep k}$ for $n\in J\_k$. So,
\begin{equation\*}
 F\_{\ep}(x)<2^\ep\sum\_{k=1}^\infty 2^{-\ep k}|J\_k|. 
\end{equation\*}
Take now any real $\de>0$ and, in view of (1), let $k\_\de$ be a natural number such that
\begin{equation\*}
 \sum\_{k\ge k\_\de}|J\_k|/k<\de. 
\end{equation\*}
Let $c\_\de:=\sum\_{k=1}^{k\_\de-1}|J\_k|$.
Then
\begin{align\*}
 F\_{\ep}(x)&<c\_\de+2^\ep\sum\_{k\ge k\_\de} k2^{-\ep k}|J\_k|/k \\ 
 & \le c\_\de+2^\ep\max\_{k\ge1}(k2^{-\ep k})\,\sum\_{k\ge k\_\de} |J\_k|/k \\ 
 & <c\_\de+2^\ep\frac1{\ep e\ln2}\,\de \\ 
 & <c\_\de+\de/\ep 
\end{align\*}
if $\ep\in(0,1/2)$.
So,
\begin{equation}
 \limsup\_{\ep\downarrow0}\frac{F\_{\ep}(x)}{1/\ep}\le\de,
\end{equation}
for every real $\de>0$. Now ($\*$) follows, in view of (0).


