$C_j$ is a sequence of matrices in $\mathbb C^{n \times n}$ and the identity $$\max_{j,k}|A_{j,k}|\leq \|A\|\leq n\max_{j,k}|A_{j,k}|$$ is known. Show that $\sum_{j=0}^\infty C_j$ converges if $\sum_{j=0}^\infty \|C_j\|$ does.
I tried a proof by contradiction:
If $\sum_{j=0}^\infty C_j$ does not converge, then for any $0\leq m <\infty$ there exists an $n>m$ such that
$$\max\Big|\sum_{j=0}^m C_j\Big|\leq\max\Big|\sum_{j=m}^n C_j\Big|$$
Now since any matrix norm satisfies $\|A+B\|\leq\|A\|+\|B\|$ we can say that
$$\Big\|\sum_{j=m}^n C_j\Big\| \leq \sum_{j=m}^n \| C_j \|$$
Because of the above identity we can write
$$\max\Big|\sum_{j=0}^m C_j\Big|\leq\max\Big|\sum_{j=m}^n C_j\Big|\leq\Big\|\sum_{j=m}^n C_j\Big\| \leq \sum_{j=m}^n \| C_j \|$$
but we can pick $m$ arbitrarily large, hence $\sum_{j=m}^n \| C_j \|$ diverges when $m\to \infty$.
So I hopefully showed that the $\|C_j\|$ series diverges if the $C_j$ series does, too, which is the "other direction" of the implication to be shown. As pointed out in the comments, this forgets the case where the series oscillates, but would the proof work out if that were taken care of?
Edit: I forgot to mention the $C_j$ are matrices, I hope that doesn't change much.