Summation manipulation. 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*}
 A: 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.
A: 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.
