Iterated series of a particular form Let $\{x_{i,j}:i\ge1,j\ge1\}$ be real numbers. In general,
$$
\sum_{i=1}^\infty\sum_{j=1}^\infty x_{i,j}\ne\sum_{j=1}^\infty\sum_{i=1}^\infty x_{i,j}.
$$
If $x_{i,j}\ge0$ for $i\ge1$ and $j\ge1$ or either of the iterated series converges absolutely, then both of the iterated series are equal (see, for example, Tonelli's theorem and this question).
I'm interested in an iterated series of a particular form.

Suppose that $x_{i,j}=a_{i,j}b_j$, where $\{a_{i,j}:i\ge1,j\ge1\}$ and $\{b_j:j\ge1\}$ are real numbers and $a_{i,j}\ge0$ for $i\ge1$ and $j\ge1$. Are the iterated series
  $$
\sum_{i=1}^\infty\sum_{j=1}^\infty a_{i,j}b_j\quad\text{and}\quad\sum_{j=1}^\infty\sum_{i=1}^\infty a_{i,j}b_j
$$
  equal if we know that $\sum_{i=1}^\infty a_{i,j}<\infty$ for $j\ge1$, $\sum_{j=1}^\infty\sum_{i=1}^\infty a_{i,j}=\infty$, but the series $\sum_{j=1}^\infty\sum_{i=1}^\infty a_{i,j}b_j$ converges (it does not converge absolutely)?

Since $\{b_j:j\ge1\}$ are not necessarily non-negative and the latter iterated series does not converge absolutely, I cannot use Tonelli's theorem. So I'm interested in any results that can be used when Tonelli's theorem is not applicable. Also, I was trying to find a counterexample, but I didn't manage to find one.
Any help is much appreciated!
 A: So from the comments, we are trying to find a counterexample to
$$\begin{cases}a_{i,j}\geq 0 \\b_j ∈ℝ \\\sum^∞_{j=1}\sum^∞_{i=1}a_{i,j}=∞ \\  \sum^∞_{j=1}\sum^∞_{i=1}|a_{i,j}b_j|=∞\\
\sum^∞_{j=1}\sum^∞_{i=1}a_{i,j}b_j<∞\\ \end{cases} \overset{?}{\implies} \sum^∞_{j=1}\sum^∞_{i=1}a_{i,j}b_j = \sum_{i=1}^∞ \sum_{j=1}^∞ a_{i,j}b_j$$
i.e. is this enough to exchange the order of summation?
One can check that my first try from the comments,
$$a = {\tiny \begin{pmatrix}2   \\ \frac{1}{2} & \frac{1}{2} \\
&\frac{1}{3}&\frac{1}{3}\\ 
&&\frac{1}{4}&\frac{1}{4}\\ 
&&&\frac{1}{5}&\frac{1}{5}\\ 
&&&&\ddots&
\end{pmatrix}},\quad b_j = (-1)^{j+1}$$
(omitted entries are $0$) satisfies the assumptions but both sums are $\log(4)$.
After much trial and error I have a counterexample :)
Let $x_n = \frac{1}{n}$,$b_n = (-1)^{n+1}$. It is known that $∑x_nb_n$ converges conditionally to $\log 2$, and we can also find some bijection $\sigma:\Bbb N→\Bbb N$ such that $\sum x_{\sigma(n)}b_{\sigma(n)}$ is any value you want  $\xi\inℝ ∪ \{± ∞\}$. We will do a similar thing to force a different value when you switch the order of summation. The idea is that (for $n>1$) we will have the $n$th row sum to $x_nb_n$, so that the sum of row sums gives $\log 2$, but each sum along a column can be controlled.
Fix a $\xi∈\Bbb R$.
Start with $b_j = (-1)^{j+1}$, and $a$ the infinite matrix of $0$s and a single $x_1=1$,
$$a=\begin{pmatrix}
1 & & &\\
& & &\\
\end{pmatrix}$$
Once we give a value to $a_{i_0,j_0}$, we will choose to ensure that the values above and under it are $0$ i.e. $\{a_{i_0,j_0± k} : k∈\Bbb N\} = \{0\}$. Thus before we finish defining $a$, we can talk of the partial sum of column sums, $$S_N := \sum_{i=1}^N\sum_{j=1}^∞ a_{ij}b_j$$
To decide which value to include next, check if $S_N>\xi$ or $X_N\leq \xi$. If $S_N>\xi$, then we want to include some negative values in the next column, i.e. we will populate the next columns with entries in even rows,
$$a=\begin{pmatrix}
1 \\
 &x_2 \\
 &0\\
 &&\ddots\\
&&&0\\
&&&x_{N_1}\\
\\
&&&&&
\end{pmatrix}$$
And we keep going until $S_{N_1}\leq\xi$. (This happens eventually as $\sum x_{2n} =∞) $. At this point, we return to the odd rows we have yet to fill in, and start filling in values in the next columns,
$$a=\begin{pmatrix}
1 \\
 &x_2 \\
 &0 &&&x_3\\
 &&\ddots&&&\ddots\\
&&&0&&&&x_{N_2}\\
&&&x_{N_1}\\
\\
&&&&&
\end{pmatrix}$$
until $S_{N_2} < \xi$.
In this way, we will fill every row with exactly one number $x_n$. Thus the row sum $\sum_ia_{ij}b_j = b_j x_j$, and so $$\sum_j\sum_ia_{ij}b_j = \sum_j b_j x_j = \log 2$$
However, by construction, the partial sum of column sums tends to $\xi$,
$$S_N = \sum_{i=1}^N\sum_{j=1}^∞ a_{ij}b_j\xrightarrow[N→∞]{}\xi$$ (There is an simple modification for $\xi=±∞$.)
