Is there an asymptotic bound between converging and diverging series? [closed]

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?

Answer 1

1. 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 $b_n/a_n\to0$.

2. 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 $b_n/a_n\to\infty$.