# Interchange of summation in double series

I have been trying to understand the reasoning behind the following change of summation but still don't get it down, could anyone explain me what is happening?

The summation is $$\sum_{n=0}^{\infty}\frac{z^n}{n!}\sum_{r=0}^n s(n,r)x^r = \sum_{r=0}^{\infty} x^r \sum_{n=r}^{\infty} s(n,r) \frac{z^n}{n!}$$

where $s(n,r)$ is the Stirling number of the first kind.

Hint: Maybe the following representation of indices is helpful

\begin{align*} \sum_{n=0}^{\infty}\sum_{r=0}^{n}a_{n,r}=\sum_{0\leq r\leq n< \infty}a_{n,r}=\sum_{r=0}^{\infty}\sum_{n=r}^\infty a_{n,r} \end{align*}

The key observation is that the original double summation can be written

$$\sum_{0\le r\le n}x^rs(n,r)\frac{z^n}{n!}\;,$$

where $r$ and $n$ are of course understood to range over integers. If we slice up the set

$$\{\langle n,r\rangle\in\Bbb N\times\Bbb N:r\le n\}$$

of allowable pairs of indices by choosing $n$ and looking at the acceptable choices for $r$, we get your lefthand side. However, we could pick $r$ first instead and ask for the acceptable choices for $n$. Clearly $r$ can be any non-negative integer, and once we’ve settled on that $r$, $n$ can be any integer that is not smaller than $r$: that gives us the righthand side. It’s entirely analogous to the manipulation

$$\int_0^\infty\int_0^xf(x,y)\,dy\,dx=\int_0^\infty\int_y^\infty f(x,y)\,dx\,dy\;,$$

except that with a formal power series we needn’t worry about convergence.

If we set

$$a_{n,r}=x^rs(n,r)\frac{z^n}{n!}\;,$$

the following diagram may be helpful.

$$\begin{array}{c|cc} n\backslash r&0&1&2&3&\ldots\\ \hline 0&a_{0,0}\\ 1&a_{1,0}&a(1,1)\\ 2&a_{2,0}&a(2,1)&a(2,2)\\ 3&a_{3,0}&a(3,1)&a(3,2)&a(3,3)\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots \end{array}$$

The lefthand side of your equality corresponds to calculating the row sums and then adding them up. The righthand side corresponds to calculating the column sums and then adding them up.