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?



Is there an asymptotic bound between converging and diverging series?
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 $b\_n/a\_n\to\infty$.


