How can this be true?
\begin{align*} \sum_{n=1}^\infty \sum_{k=n}^\infty \frac 1{k^3} &= \sum_{k=1}^\infty \sum_{n=1}^{k} \frac 1{k^3}\\ \end{align*}
How can this be true?
\begin{align*} \sum_{n=1}^\infty \sum_{k=n}^\infty \frac 1{k^3} &= \sum_{k=1}^\infty \sum_{n=1}^{k} \frac 1{k^3}\\ \end{align*}
On the left-hand side, imagine the following infinite sum of infinite sums:
$\begin{aligned} \frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots\\ \frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots\\ \frac{1}{3^3}+\frac{1}{4^3}+\cdots\\ \frac{1}{4^3}+\cdots \end{aligned}$
This is equal to the infinite sum $\frac{1}{1^3}+\frac{2}{2^3}+\frac{3}{3^3}+\frac{4}{4^3}+\cdots$ or $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots$
Now look at the right-hand side. The expression $\frac{1}{k^3}$ does not depend on $n$ at all, so the inner sum is just $\frac{k}{k^3}$, and $\sum_{k=1}^\infty \frac{k}{k^3} = \sum_{k=1}^\infty \frac{1}{k^2}$
which is the same as the left-hand side, as we showed above.
Let $a_{n,k}:=\frac 1{k^3}\chi_{\{(n,k),n\leq k\}}\geq 0$. The first sum is $\sum_{n=1}^{+\infty}\sum_{k=1}^{+\infty}a_{n,k}$ and the second one is $\sum_{k=1}^{+\infty}\sum_{n=1}^{+\infty}a_{n,k}.$ Switching the two sums is allowed as we deal with non-negative terms.