Is asymptotic growth bound on a sequence equivalent to an asymptotic growth bound on its partial sum? The following question was asked at https://mathoverflow.net/questions/360053/asymptotic-growth-bound-on-a-sequence-equivalent-to-an-asymptotic-growth-bound-o, but then deleted by the user:

I posted the following question on [Math Stack Exchange][1] on April 2, where it's been unanswered for over a month. I hope the good denizens of Overflow will allow me to migrate it here in hope of an answer or some advice on where to look further.
Question:
Let $(a_n)$ and $(b_n)$ be sequences of positive real numbers. Denote by $o$ the "little-oh" Landau symbol. Is it possible, in general, to find a third sequence $(c_n)$ such that $\sum_{k=0}^n a_k = o(b_n)$ if and only if $a_n = o(c_n)$? Is there a formula for such a $(c_n)$ in terms of $(a_n)$ and $(b_n)$?
It's also possible that some generalization of classical asymptotic analysis can be used to obtain a result in the same spirit, and I'm interested in hearing if this is so.
[1]: https://math.stackexchange.com/questions/3606416/asymptotic-growth-bound-on-a-sequence-equivalent-to-an-asymptotic-growth-bound-o

I think this question may make sense. Below is the answer to it. 
 A: The answer is yes or no, depending on how the quantifiers are placed. 
If the question is this: Is it true that 

$\forall (a_n)\ \forall (b_n)\ \exists (c_n)\ \ \big[\sum_{k=0}^n a_k=o(b_n)\iff a_n = o(c_n)\big]$?

then the answer is yes. Everywhere here, $(a_n),(b_n),(c_n)$ are sequences of positive real numbers.
Indeed, for any such $(a_n)$ and $(b_n)$, for all $n$ just let $c_n:=na_n$ if $\sum_{k=0}^n a_k=o(b_n)$ and $c_n:=a_n$ otherwise. 

More interesting is this question:  Is it true that 

$\forall (b_n)\ \exists (c_n)\ \forall (a_n)\ \ \big[\sum_{k=0}^n a_k=o(b_n)\iff a_n=o(c_n)\big]$?

Here the answer is no. Indeed, suppose that, to the contrary, the latter highlighted statement holds. For all natural $n$ let 
$$b_n:=n\ln n.$$
Then I claim that for any $(c_n)$ such that 
\begin{equation*}
 \forall (a_n)\ \ \big[\sum_{k=0}^n a_k=o(b_n)\iff a_n=o(c_n)\big] \tag{0}
\end{equation*}
we have 
\begin{equation*}
 c_n>\sqrt n
\end{equation*}
eventually, that is, for all large enough $n$. 
Indeed, suppose otherwise: that for some natural $n_1<n_2<\cdots$ and all natural $j$
\begin{equation*}
 c_{n_j}\le\sqrt{n_j}. \tag{1}
\end{equation*}
For all natural $k$, let now 
\begin{equation*}
 a_k:=\sum_{i=1}^\infty\sqrt{n_{i^2}}\,I\{k=n_{i^2}\},
\end{equation*}
where $I\{\cdot\}$ denotes the indicator. That is, $a_k=\sqrt{n_{i^2}}$ if $k=n_{i^2}$ for some natural $i$, and $a_k=0$ otherwise. Then
\begin{equation*}
 \sum_{k=0}^n a_k
 =\sum_{i=1}^\infty\sqrt{n_{i^2}}\,I\{n_{i^2}\le n\}
 \le\sum_{i=1}^\infty\sqrt n\,I\{i^2\le n\}\le n=o(b_n). 
\end{equation*}
So, by (0), $a_n=o(c_n)$, whence 
\begin{equation*}
 \sqrt{n_{i^2}}=a_{n_{i^2}}=o(c_{n_{i^2}})=o(\sqrt{n_{i^2}}),
\end{equation*}
by (1). This contradiction proves that indeed $c_n>\sqrt n$ eventually. 
So, letting now $a_n:=\sqrt n/\ln(n+1)$, we have the condition $a_n=o(c_n)$ satisfied. However, here 
\begin{equation*}
 \sum_{k=0}^n a_k\sim\tfrac23\, n^{3/2}/\ln(n+1),
\end{equation*}
which is not $o(b_n)$. Thus, the second highlighted statement is false. 
