Is there an asymptotic bound between converging and diverging series?

Question

Let us define for every $k\in\mathbb{N}$ and every large enough $x\in \mathbb{R}$ , $\log^{[k]}(x) = \begin{cases} \log^{[k-1]}(\log(x)) & k>0 \\ x & k=0 \end{cases}.$ It is well known, from the series condensation theorem, that for $0< p\in\mathbb{R}, k\in\mathbb{N}$ and large enough $M\in\mathbb{N}$ that $\sum_{n=M}^\infty a_n(p,k) = \sum_{n=M}^\infty\frac 1{n\log(n)\log^{[2]}(n)\ldots\log^{[k-1]}(n)(\log^{[k]}(n))^p}$ diverges for $p\leq1$ and converges for $p>1$ .
By enlarging $k$ to $k+1$ for a fixed $0 < p\leq1$ we see that there is no asymptotically lower bound of the general term of these diverging series (we can always find an asymptotically smaller general term that still diverges for these series), i.e. $\begin{split} \lim_{n\rightarrow\infty}\frac{a_n(p,k+1)}{a_n(p,k)} & = \lim_{n\rightarrow\infty}\frac{{n\log(n)\log^{[2]}(n)\ldots\log^{[k]}(n)^p}}{{n\log(n)\log^{[2]}(n)\ldots\log^{[k]}(n)(\log^{[k+1]}(n))^p}} \\ &= \lim_{n\rightarrow\infty}\frac{1}{{{(\log^{[k]}(n))^{1-p}}(\log^{[k+1]}(n))^p}} \\ & \leq \lim_{n\rightarrow\infty}\frac{1}{{(\log^{[k+1]}(n))^p}} = 0 \end{split}$

and $\sum a_n(p,k+1)$ is still diverging.
In addition by reducing $p>1$ to $p-\varepsilon>1$ by a small enough $\varepsilon>0$ for a fixed $k$ we see that there is not asymptotically upper bound of the general term of these converging series (we can always find an asymptotically larger general term that still converges for these series), i.e. $\begin{split} \lim_{n\rightarrow\infty}\frac{a_n(p,k)}{a_n(p-\varepsilon,k)} & = \lim_{n\rightarrow\infty}\frac{{n\log(n)\log^{[2]}(n)\ldots\log^{[k]}(n)^{p-\varepsilon}}}{{n\log(n)\log^{[2]}(n)\ldots(\log^{[k]}(n))^{p}}}\\ & = \lim_{n\rightarrow\infty}\frac{1}{{(\log^{[k+1]}(n))^\varepsilon}} = 0 \end{split}$

and $\sum a_n(p-\varepsilon,k)$ is still converging.

And now the questions:

Is it true in general?
Does for all diverging series $\sum a_n$ there exists a diverging series $\sum b_n$ such that $\lim_{n\rightarrow\infty}\frac{b_n}{a_n} = 0\;?$
And does for all converging series $\sum a_n$ there exists a converging series $\sum b_n$ such that $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=0\;?$
Finally, is there an asymptotic bound between converging and diverging series?

Yes, this works. For example in the first part, given a divergent series, take $b_n=1$ for a while, then $b_n=a_n/2$ etc. and do this gingerly (in other words, wait long enough until decreasing the factor, for example, make sure that each segment has sum $\ge 1$ ). — Christian Remling, May 23 '21 at 23:09
Thanks @ChristianRemling Will try to take your approach to find also the converging part.
My final question is on some asymptotic bound between converging and diverging series. — Niv Sarig, May 23 '21 at 23:22
Here is a very good answer to my question https://mathoverflow.net/questions/49415/nonexistence-of-boundary-between-convergent-and-divergent-series?rq=1 — Niv Sarig, May 23 '21 at 23:43
I’m voting to close this question because it is a duplicate. — Mark Wildon, May 24 '21 at 10:49
Does this answer your question? Nonexistence of boundary between convergent and divergent series? — მამუკა ჯიბლაძე, May 29 '21 at 05:05

Iosif Pinelis · Accepted Answer · 2021-05-23T23:28:03.953

Suppose $a_n>0$ for all $n$ and $\sum_n a_n=\infty$ . Let $s_n:=\sum_{j\le n}a_j$ . Let $n_1<n_2<\cdots$ be natural numbers such that $s_{n_{k+1}}-s_{n_k}>k$ for all $k$ . Let $b_n:=a_n/k$ if $n_k<n\le n_{k+1}$ , so that $t_{n_{k+1}}-t_{n_k}>1$ for all $k$ , where $t_n:=\sum_{j\le n}b_j$ . Then $\sum_n b_n\ge\sum_k 1=\infty$ and $b_n/a_n\to0$ .
Suppose $a_n>0$ for all $n$ and $s:=\sum_n a_n<\infty$ . Let again $s_n:=\sum_{j\le n}a_j$ . Let $n_1<n_2<\cdots$ be natural numbers such that $s_{n_{k+1}}-s_{n_k}\le s-s_{n_k}<1/k^3$ for all $k$ . Let $b_n:=k a_n$ if $n_k<n\le n_{k+1}$ , so that $t_{n_{k+1}}-t_{n_k}<1/k^2$ for all $k$ , where again $t_n:=\sum_{j\le n}b_j$ . Then $\sum_n b_n\le\sum_k 1/k^2<\infty$ and $b_n/a_n\to\infty$ .

Thanks Iosif. This is pretty similar to Christian's answer. I believe the converging part should be somehow similar... — Niv Sarig, May 23 '21 at 23:26
And here is some more on this topic :-) https://mathoverflow.net/a/49447/237946 — Niv Sarig, May 23 '21 at 23:48

Is there an asymptotic bound between converging and diverging series?

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