Double summation of infinite series convergence I am currently reading Stephen Abbot analysis. I am also solving problems in it. I encountered the following problems that I am a little bit confused about. Exercise 2.8.3 (b). I am trying to argue that $s_{nn}$ is cauchy. What I am confused about is that don't we by the reverse triangle inequality that we have $| s_{nn} - s_{mm} | \geq | t_{nn} - t_{mm} |$ ? Is there something that I am missing ? 

 A: Note that by the usual triangle inequality
$$|s_{mm} - s_{nn}| = \left|\sum_{j=n+1}^m \sum_{k=n+1}^m a_{jk} + \sum_{j=1}^n \sum_{k=n+1}^m a_{jk} + \sum_{j=n+1}^m \sum_{k=1}^n a_{jk}\right| \\ \leqslant \sum_{j=n+1}^m \sum_{k=n+1}^m |a_{jk}| + \sum_{j=1}^n \sum_{k=n+1}^m |a_{jk}| + \sum_{j=n+1}^m \sum_{k=1}^n |a_{jk}|\\= t_{mm} - t_{nn} = |t_{mm} - t_{nn}|$$
A: When manipulating the double sums, it is tempting to treat them like we do with normal sums. For instance, if $n > m$, it is tempting to write
$$
s_{nn}-s_{mm} = \sum_{i=1}^n\sum_{j=1}^na_{ij}-\sum_{i=1}^m\sum_{j=1}^ma_{ij} = \sum_{i=m+1}^n\sum_{j=m+1}^na_{ij}, \quad\leftarrow\text{wrong!}
$$
and this is what I had written in my original answer about 4 years ago. However, this is not the right way to manipulate the double sums. There are still the "off-diagonal" terms to consider, which were included in the other answer by @RRL. For visualization purposes, it's helpful to draw a large square, and divide it into sections to see which parts cancel in the difference of $s_{nn}-s_{mm}$:
In particular,
$$
s_{nn}-s_{mm} = \sum_{i=m+1}^n\sum_{j=m+1}^na_{ij}+\sum_{i=m+1}^n\sum_{j=1}^ma_{ij} + \sum_{i=1}^m\sum_{j=m+1}^na_{ij}
$$
and
$$
t_{nn}-t_{mm} = \sum_{i=m+1}^n\sum_{j=m+1}^n|a_{ij}|+\sum_{i=m+1}^n\sum_{j=1}^m|a_{ij}| + \sum_{i=1}^m\sum_{j=m+1}^n|a_{ij}|.
$$
This observation, together with the triangle inequality, shows
$$
|s_{nn}-s_{mm}|\le t_{nn}-t_{mm},
$$
which proves that $s_{nn}$ is Cauchy since we assume $t_{nn}$ is Cauchy.
