Timeline for Is the harmonic series worse than any summable series?

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Mar 21, 2021 at 11:27	comment	added	Sascha		@IosifPinelis thanks a lot for your answer. I started wondering about the following, which I understand might have to be a separate question, but perhaps there is an easy work-around: Take $x_n=(1/n^2)$ and $z_n=(n). $ Now, define $F_{\varepsilon}(y) = \sum_n 2^{-\varepsilon y_n}.$ Is it true that among all $y=(y_n)$ positive such that $\sum_n x_n y_n$ is finite, we have $\liminf_{\varepsilon \to 0} \frac{F_{\varepsilon}(y)}{F_{\varepsilon}(z)} >0$?
Mar 18, 2021 at 18:13	vote	accept	Sascha
Mar 18, 2021 at 18:05	comment	added	Iosif Pinelis		@WillieWong : Good point!
Mar 18, 2021 at 17:18	comment	added	Willie Wong		Incidentally: if it were the case that the $\limsup$ of the ratio is bounded by 1 for ALL absolutely summable sequences $(x_n)$, then necessary the limit is 0 for all such sequences. This is because given a sequence $(x_n)$, if you form $(y_n)$ by listing each term of $(x_n)$ twice, then $y$ is still absolutely summable and $F_\epsilon(y) = 2 F_\epsilon(x)$. Hence it is in fact necessary to prove the stronger statement.
Mar 18, 2021 at 14:27	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 63 characters in body
Mar 18, 2021 at 14:22	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 63 characters in body
Mar 18, 2021 at 13:58	history	answered	Iosif Pinelis	CC BY-SA 4.0