This question shows a special case of an infinite double sum where rearranging the order of summation preserves the value of the expression*
$$ \sum_{i=1}^{\infty}\sum_{j=(i)}^{\infty}f(i,j)=\sum_{j=1}^{\infty}\sum_{i=1}^{j}f(i,j) $$
Or, written using an Iverson bracket
$$ \sum_{i \, \in \, \mathbb{N}} \sum_{j \, \in \, \mathbb{N}} [\,j \ge i\,] f(i,j) \;=\; \sum_{j \, \in \, \mathbb{N}} \sum_{i \, \in \, \mathbb{N}} [\,i \le j\,] f(i,j) $$
This suggests something interesting, namely that the expression $[i \le j] f(i,j)$ is special and can be summed in either order.
I'm not sure why that's the case, but I'm wondering if there are simple rules that can identify some of the expressions that are summable in either order or have the same divergence behavior in either order.
* One of the answers to the linked question justifies the identity by noting that both sides are equivalent to $$ \sum_{1 \le i \le j \le \infty} f(i,j) $$
It is not clear to me why this expression is well-defined unless the series is absolutely convergent or divergent.
Best Answer
It's not true. For example, consider $$\eqalign{f(i,i) &= 1\cr f(i,i+1) &= -1\cr f(i,j) &= 0\ \text{ otherwise}\cr}$$ Then for all $i$, $\sum_{j=i}^\infty f(i,j) = 0$ so $$\sum_{i=1}^\infty \sum_{j=i}^\infty f(i,j) = 0$$ On the other hand, $\sum_{i=1}^1 f(i,1)=1$ while $\sum_{i=1}^j f(i,j) = 0$ for $j \ge 2$, so $$\sum_{j=1}^\infty \sum_{i=1}^j f(i,j) = 1$$