When does the fact that $\underset{(k,k)}{\sum\sum} a_{k,k}$ converges imply convergence for $\underset{(k,\ell)}{\sum\sum} a_{k,\ell}$? I have come to a problem in a multivariate calculus book that I'm having some trouble with.
The book is "A Course in Multivariate Calculus and Analysis" by Ghorpade and Limaye.
The problem goes :

Let $\underset{(k,\ell)}{\sum\sum} a_{k,\ell}$ be a double series whose terms are schematically given by :
\begin{equation}
\begin{matrix}
            1 &  2   &  4   &  8 & \dots \\
 -\frac{1}{2} & -1    & -2   & -4 & \dots \\
 -\frac{1}{4} & -\frac{1}{2}   & -1   & -2 & \dots \\
 -\frac{1}{8} & -\frac{1}{4}   & -\frac{1}{2}  & -1 & \dots \\
\vdots & \vdots & \vdots & \vdots & \; 
\end{matrix}
\end{equation}
and let $A_{m,n}$ denote its $(m,n)$th partial double sum. Show that each row-series is divergent, but each column-series converges to $0$. Also, show that
$A_{m,m} \rightarrow 2$ as $m \rightarrow \infty$. Is $\underset{(k,\ell)}{\sum\sum} a_{k,\ell}$ convergent ?

I am able to solve all of the problem except for the last part, namely determining
if $\underset{(k,\ell)}{\sum\sum} a_{k,\ell}$ is convergent. My question really is under what conditions for $(a_{k,\ell})$ does :
\begin{equation}
(A_{m,m}) \text{ is convergent } \Rightarrow (A_{m,n}) \text{ is convergent }
\end{equation}
Could someone maybe provide an example where the above implication is not true ?

To be clear, we say that the double series $\sum_{(k,l)}a_{kl}$ converges to $A$ and write
$$\lim_{m,n \to \infty} A_{mn} = \lim_{m,n \to \infty}\sum_{k=1}^m\sum_{l=1}^n a_{kl} = A$$
if for any $\epsilon >0$ there is $N \in \mathbb{N}$ such that $|S_{mn} -A|< \epsilon\,$ for all $m,n > N$.  Consequently, we have $|A_{mm} - A| < \epsilon$ for all $m > N$, and it follows that if the double series converges to $A$ then we must have
$$\lim_{m \to \infty} A_{mm} = A$$
In this case, we have $\lim_{m \to \infty}A_{mm} =2$ as you have shown, and if the double series $\sum_{(k,l)}a_{kl}$ converges it must converge to $A = 2\,$ as well.
However, we can prove here that the double series does not converge using the following theorem.

If the double series converges with $\lim_{m,n \to \infty}A_{mn} = A$
and, for each $l \in \mathbb{N}$, we have convergence of the column
sum $\sum_{k=1}^m a_{kl} \to b_l$  as $m \to \infty$, then the
iterated sum converges as
$$\tag{*}\sum_{l=1}^\infty \left(\sum_{k=1}^\infty a_{kl} \right)=\lim_{n \to
 \infty}\sum_{l=1}^n b_l = A$$

As already shown for this problem, each column sum converges to $0$ and, hence, the iterated sum in (*) must also converge to $0$.  Since we cannot have both $A=0$ (the limit of the iterated sum) and $A= 2$ (the limit of $A_{mm}$), it follows that the double series does not converge.

The theorem giving the result (*) is proved by noting that there exist $N \in \mathbb{N}$ such that for all $m,n > N$ we have
$$|A_{mn} - A| < \epsilon$$
Thus, for all $n > N$,
$$\left|\sum_{l=1}^n b_l -A\right| = \left|\sum_{l=1}^n \sum_{k=1}^\infty a _{kl} \right|=\left|\lim_{m \to \infty}\sum_{l=1}^n \sum_{k=1}^m a _{kl} \right|=\lim_{n \to \infty}|A_{mn} -A| < \epsilon$$
