A kind of Cauchy's Condensation Test? Let $f:[0,\infty) \to \mathbb{R}$ be a positive , monotone decreasing function and
$ f(x) \to 0$ as $x\to \infty$
Let $\{a_n\}$ be a bounded sequence such that $a_n \ge 0$ and $\displaystyle\sum _{n=1}^{\infty} a_n=\infty$ .
Then prove that $\sum_{n\ge 1} f(n)$ and $\sum_{n\ge 1}a_nf(a_1+a_2+\cdots+a_n)$ converge or diverge together.
My attempt
The zero terms of $\{a_n\}$ contribute nothing to the convergence/divergence of
$$\sum_{n=1}^{\infty}a_n$$
and $$\sum_{n\ge 1}a_nf(a_1+a_2+\cdots+a_n).$$
So without loss of generality , let us assume that $a_n \gt 0, \forall n$.  Since $a_n$ is bounded, we must have
$m_*\le a_n \le M^*$ , for some $m_*, M^*\gt 0$
So, we have
$$a_1+a_2+\cdots+a_n \ge nm_*$$
If $m_* \ge 1$, then
$$a_1+a_2+\cdots+a_n \ge n$$
$\Rightarrow f(a_1+a_2+\cdots+a_n) \le f(n)$, since $f(x)$ is monotone decreasing function.
So by Comparision test if $\sum f(n)$ converges, then $\sum_{n\ge 1}f(\sum _{i=1}^{n}a_i)$ converges and so does $\sum_{n\ge 1}a_nf(\sum _{i=1}^{n}a_i)$ since
$$a_nf\left(\sum _{i=1}^{n}a_i\right) \le M^*f\left(\sum _{i=1}^{n}a_i\right).$$
The case for $m_* \ge 1$ ends here.
Let $M^* \le 1.$
Then $a_1+a_2+\cdots+a_n \le nM^*\le n$
$$\Rightarrow f\left(\sum _{i=1}^{n}a_i\right)\ge f(n)$$
Then by Comparision Test, if $\sum_{n\ge 1}f(n)$ diverges , then $\sum_{n\ge 1}f(\sum _{i=1}^{n}a_i)$ diverges and so does $\sum_{n\ge 1}a_nf(\sum _{i=1}^{n}a_i)$ since
$$a_nf\left(\sum _{i=1}^{n}a_i\right) \ge m_*f\left(\sum _{i=1}^{n}a_i\right).$$
The case for $M^* \le 1$ ends here.
I do not know how to approach for $m_* \le 1 \le M^*$ and moreover I am familiar with the missing reverse implications in the above cases.
Do you have any ideas? Thanks for your time.
As a sidenote, this is a previously asked question on MSE but the question was voluntarily removed by the OP for some unknown reasons (at least unknown to me). I would really like the question to be answered.
 A: PARTIAL answer:
$(a_n)$ is bounded so there exists an $M > 0$ such that for all $n$ we have $0 \leq a_n \leq M$. Moreover, as you suggested, we can assume without loss of generality that $a_n > 0$ for all $n$.
Let $N_k = \min \{ n : a_1 + \dots + a_n \geq k\}$.
$N_k$ is well defined and finite for all $k$ because $\sum_n a_n = \infty$.
Now since all terms are non-negative we can write
$$
\sum_{n \geq 1} a_n f(a_1 + \dots + a_n)
= \sum_{k\geq 0}  \sum_{n=N_k}^{N_{k+1}-1} a_n \underbrace{f(a_1 + \dots + a_n)}_{\leq f(k)}
\leq \sum_{k \geq 0}  \underbrace{\left(\sum_{n=N_k}^{N_{k+1}-1} a_n\right)}
_{\leq a_{N_k} + 1 \leq M + 1} f(k)
\leq (M + 1) \sum_{k \geq 0} f(k)
$$
where the sum in parenthesis in the center may be empty and thus equal to zero.
This proves that the divergence of $\sum_n a_n f(a_1 + \dots + a_n)$ implies the divergence of $\sum_n f(n)$.
Now we write similarly
$$
\begin{aligned}
\sum_{n \geq 1} a_n f(a_1 + \dots + a_n)
&= \sum_{k \geq 0}  \sum_{n=N_k}^{N_{k+1}-1} a_n \underbrace{f(a_1 + \dots + a_n)}_{\geq f(k+1)} \\
&\geq \sum_{k \geq 0}  \left(\sum_{n=N_k}^{N_{k+1}-1} a_n\right) f(k+1) \\
&= \sum_{k \geq 1}  \underbrace{\left(\sum_{n=N_{k-1}}^{N_k-1} a_n\right)}_{b_k} f(k)
= \sum_{k \geq 1}  b_k f(k).
\end{aligned}
$$
Notice that $ 0 \leq b_k \leq M + 1$ for all $k$, and that $\sum_k b_k = \sum_n a_n = \infty$.
In fact, we even have that for all $K$,
$$
\max_{\substack{0 \leq k < K \\ N_k < N_K}} k \ \leq \sum_{k=1}^K b_k = \sum_{n=1}^{N_K-1} a_n \leq K
$$
The idea here is that since $f$ is decreasing, you can always 'shift' the weights towards bigger values of $k$ to get a lower bound of $\sum_k b_k f(k)$. If you get a lower bound of the form $ C \sum_{k \geq N} f(k)$ with $N$ an integer and $C$ a constant you are done because this means that the divergence of $\sum_n f(n)$ implies the divergence of $\sum_n a_n f(a_1 + \dots + a_n)$. I believe that finding such a lower bound is doable.
There might be a nicer solution though...
Hope this helps a bit even though it is not a full solution!
A: I think I have found a solution. Here it goes .
Proof :-
Since $f$ is monotone , it is integrable on every bounded interval.
As  I already said, let $a_n \gt 0$
Let $M=\sup \{a_n : n\in \mathbb{N} \}$
Let $s_n=a_1+a_2+..+a_n,  n\in \mathbb{N} $
Therefore $s_n \to \infty$   as $n\to \infty$
Consider the partition $P$ of $[0, s_n]$  given by
$\{0, s_1, s_2,..., s_n\}$
Since $f(x)$ is decreasing , the lower sum
$L(P,f)=a_1f(a_1)+a_2f(a_1+a_2)+...+a_nf(a_1+a_2+...a_n)$.
$=\sum_{i=1}^n a_i f (s_i)$
So $L(P,f)$  is one of our required sum here.
We know that
$L(P,f) \le \int_0^{s_n}f(x)dx \quad (1)$
Similarly  $U(P,f)$  is the Upper Reimann  Sum and it can be easily proved
$U(P,f)-L(P,f) \le ||P|| \{f(0)-f(s_n)\}$
$\le Mf(0)=k$   (say), a constant
$L(P,f) \ge U(P,f)-k \ge \int_0^{s_n} f(x)dx-k $
Therefore from $(1)$
$\int_0^{s_n}f(x)dx -k \le L(P,f) \le \int_0^{s_n} f(x)dx $
and thus
$\int_0^{s_n}f(x)dx -k \le\sum_{i=1}^n a_i f (s_i) \le \int_0^{s_n} f(x)dx \quad (2) $
Now, let $F(x)=\int_0^x f(t) dt$.
Then $F(x)$ is monotone increasing since $f$ is positive.
Let $\lim _{n\to \infty} \sum_{i=1}^n a_i f (s_i)$   exist .
Then clearly, $\lim_{n\to \infty} F(s_n)$  exist from $(2)$. Now we must have
$\lim_{x\to \infty} F(x)\lt \infty$
otherwise $\lim_{x\to \infty} F(x)=\infty$   and hence $\lim_{n\to \infty} F(s_n)=\infty$,    a contradiction.
So  $\lim_{x\to \infty}F(x) \lt \infty \Rightarrow \lim_{n\to \infty} F(n)\lt \infty$
and hence $\sum_{n\ge 1} f(n)$   exist by Abels Integral Test
Thus if $\sum_{n\ge 1}a_nf(\sum_{i=1}^na_i)$ exist then $\sum_{n\ge 1}f(n)$ also exist.
Arguing in the reverse way, the converse can be shown.
