Take any $(x_{kl})\in \mathbb{R}^{\infty\times\infty}$ such that $\sum_{j=1}^\infty x_{kj}$ converges for each k and $(x_{1l},x_{2l},\dots)$ converges for each $l$. If there exists a real sequence $(K_1,K_2,\dots)$ such that $|x_{kl}|\leq K_l$ for each $l$,and $\sum^\infty K_i$ converges, then
$$
\lim_{k\to\infty}\sum_{j=1}^\infty x_{kj}=\sum_{j=1}^{\infty}\lim_{k\to\infty}x_{kj}
$$
The only things I observe are that for any $m$, $x_{k1}+\dots+x_{km}\leq|x_{k1}|+\dots+|x_{km}|\leq K_1+\dots+K_m$, $\lim_{m\to\infty}x_{km}=0$. Except these I have no clue where to start. Many thanks for any help!
Best Answer
Define $\lim_{k\to\infty}x_{kl}=x_l$. Since $|x_{kl}|\leq K_l$ for each $l$, then $|x_l|\leq K_l$. Thus, $\sum x_l$ is absolutely convergent by comparison test. This implies RHS limit exists.
Now given $\varepsilon>0$ choose $m$ large enough so that, $\sum_{j=m+1}K_j<\frac{\varepsilon}{3}$. From, $|x_{kl}-x_l|\leq |x_{kl}|+|x_l|\leq2K_l$ we have
\begin{gather} |\sum_{j=1}^\infty x_{kj}-\sum_{j=1}^\infty x_j|=|\sum_{j=1}^m(x_{kj}-x_j)+\sum_{j=m+1}^\infty(x_{kj}-x_j)|\\ \leq \sum_{j=1}^m |x_{kj}-x_j|+ \sum_{j=m+1}^\infty 2K_j\\ < \sum_{j=1}^m |x_{kj}-x_j|+\frac{\varepsilon}{2} \end{gather}
Since for each $l$, $\lim_{k\to\infty}x_{kl}=x_l$, we can choose the first sum arbitrarily small and the theorem follows.