What's the easiest, most concise, way to prove this simple swapping of order of nested summation? $$ \sum\limits_{n=1}^{N} \, \sum\limits_{k=1}^{n} \, a_{nk} \ = \ \sum\limits_{k=1}^{N} \, \sum\limits_{n=1}^{k} \, a_{nk} $$
for $N \in \mathbb{N} < \infty$
 A: Swapping the order of finite summations can't change the value; only for infinite sums does addition become non-commutative. However, you're not just swapping the order, you're swapping the limits. If you imagine arranging the elements in an $N \times N$ array, with the first index going from top to bottom, and the second going from left to right, then the LHS sums the bottom left triangle, while the RHS sums the top right (with the diagonal being included in both). 
A similar thing happens if you do a double integral of $xy^2$. If you integrate $y$ from 0 to $x$, you get $\frac{x^4}3$. Integrating that from 0 to $t$ gives you $\frac{t^5}{15}$. If you instead integrated $x$ from 0 to $y$, you get $\frac{y^4}2$, and integrating that from 0 to $t$ gives you $\frac{t^5}{10}$.
A: One way to prove the identity
$$\sum\limits_{n=1}^{N} \, \sum\limits_{k=1}^{n} \, a_{nk} \ = \ \sum\limits_{k=1}^{N} \, \sum\limits_{n=k}^{N} \, a_{nk}
$$
is to introduce indicator notation. Use the expression $I(a \le b)$ to represent the value $1$ when $a\le b$ and value $0$ otherwise:
$$
I(a \le b) \triangleq \begin{cases}
1 \qquad & a \le b \\
\\
0 \qquad & a > b \\
\end{cases}
$$
Then
$$
\sum\limits_{n=1}^{N} \, \sum\limits_{k=1}^{n} \, a_{nk}
\stackrel{(1)}=
\sum\limits_{n=1}^{N} \, \sum\limits_{k=1}^N \, a_{nk} \ I(k\le n)
\stackrel{(2)}=\sum\limits_{k=1}^N \sum\limits_{n=1}^{N}\, a_{nk} \ I(k\le n)
\stackrel{(3)}=\sum\limits_{k=1}^N \sum\limits_{n=k}^{N}\, a_{nk}\,.
$$
In step (1) we multiply $a_{nk}$ by an indicator to allow the inner sum to range over all $k=1,\ldots,N$. Now both sums range from $1$ to $N$, so we can swap the order of summation, which is step (2). Step (3) then removes the indicator by absorbing it into the limits of the inner sum.
A: It is sometimes useful to represent the index range of summation as inequality chain. This way the region of validity can be often seen easily.

We obtain
  \begin{align*}
\sum_{n=1}^N\sum_{k=1}^na_{nk}=\sum_{\color{blue}{1\leq k\leq n\leq N}}a_{nk}=\sum_{k=1}^N\sum_{n=k}^Na_{nk}
\end{align*}

