Let $(x\_n)$ be a monotonically decreasing sequence of positive real numbers that is also summable.


Let $(y\_n)$ be a sequence of positive real numbers such that
$\sum\_n x\_n y\_n$ converges.


Let $(z\_n)$ be a monotonically increasing sequence of positive real numbers such that $\sum\_n x\_n z\_n =\infty.$


Assume that the sequences $y\_n$ and $z\_n$ are such that $2^{-\varepsilon y\_n}$ and $2^{-\varepsilon z\_n}$ are summable for every $\varepsilon>0.$ Does it follow that there is some $\delta>0$ such that


$$ \sum\_n \Big(2^{-\varepsilon y\_n}-2^{-\varepsilon z\_n}\Big) \ge 0 \text{ for all } \varepsilon \in (0,\delta)?$$


The motivation for this statement to be true is that $z\_n$ should be larger most of the time than $y\_n$ and we capture this most of the time by taking $\varepsilon$ small.


Please let me know if you have any comments, questions or remarks.



Comparing divergent and convergent sums
No. For example $x\_n=2^{-n}$, $y\_n=n$, and we now keep $z\_n$ constant on long intervals. More precisely, consider first $\epsilon=1$, and set $z\_1= \ldots = z\_{N\_1}=c\_1$, with $N\_1$ taken so large that $N\_1 2^{-1\cdot 1}>\sum 2^{-1\cdot n}$.


We then continue in the same way: let $z\_{N\_1+1}=\ldots = z\_{N\_2}= c\_2$, and again take $N\_2$ so large that the inequality you want fails for $\epsilon=1/2$ etc.


Here we must choose the $c\_n$ such that $\sum 2^{-n}z\_n=\infty$ and $\sum 2^{-\epsilon z\_n}<\infty$. I'll take $c\_n=2^{N\_{n-1}}$ to guarantee the first property. If the $n$th step deals with $\epsilon=1/n$, then, since $\sum n2^{-\epsilon n} \simeq 1/\epsilon^2$, I can take $N\_n=n^3 2^{N\_{n-1}/n}$ to make sure your inequality fails.


With these choices in place, the final condition becomes
$$
\sum N\_n 2^{-\epsilon 2^{N\_{n-1}}} =\sum n^3 2^{(-\epsilon+1/n)2^{N\_{n-1}}}< \infty
$$
for all $\epsilon >0$, and this holds because the $N\_n$ increase rapidly.


