Comparing divergent and convergent sums

Question

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.

Answer 1

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.