# Comparing divergent and convergent sums

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.

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.
• thanks a lot for this wonderfully worked out example. If I understand things correctly, then what matters here of course is that I am looking for this delta neighbourhood. I could have also asked if $\liminf_{\epsilon \to 0} \frac{\sum_n 2^{-\varepsilon y_n}}{\sum_n 2^{-\varepsilon z_n}}>0$ holds true to better quantify their relative size to one another? I am not entirely sure how the series of your constructed $z_n$ scales with $\varepsilon$ so I don't see it right away? Mar 20, 2021 at 19:11
• @PritamBemis: I think the basic issue here is that $\sum 2^{-\epsilon u_n}$ (for $u=y$ or $u=z$) only depends on what values $u_n$ takes, not on where exactly these occur, while the other series $\sum x_nu_n$ can of course be quite sensitive to rearrangements (for given $x$). This suggests that we shouldn't really expect a very close connection between the two, and I think the situation is still essentially the same in the modification you suggested. Mar 20, 2021 at 20:41