Suppose we have the following function of the natural numbers $k \in {\mathbb N}$:
$$f(k) = \sum_{r=0}^k \sum_{h=0}^{r} (h+r)!$$
We would like to extend $f$ to all of ${\mathbb R}^+$. We know that $(h+r)! \approx \Gamma(h+r+1)$, but what is the right way to "continuize" the discrete summations?
Is there a general strategy?
In other words, suppose we have function $q:{\mathbb N \times \mathbb N} \rightarrow \mathbb N$ and an extension $q^*:{\mathbb R^+ \times \mathbb R^+} \rightarrow \mathbb R$ i.e. $q^* |_{\mathbb N \times \mathbb N} \equiv q$. Now given a function $$p(k) = \sum_{r=0}^k \sum_{h=0}^{r} q(h,r)$$ How could one extend $p$ to a smooth differentiable function $p^*$ that is defined on all of ${\mathbb R}^+$? Let's assume that $q^*$ is continuous, infinitely differentiable and any other properties needed.
My first step would be to reduce the number of summations.
$\begin{array}\\ f(k) &= \sum_{r=0}^k \sum_{h=0}^{r} (h+r)!\\ &= \sum_{r=0}^k \sum_{h=r}^{2r} h!\\ &= \sum_{h=0}^{2k} \sum_{r=\lceil h/2 \rceil}^h h!\\ &= \sum_{h=0}^{2k} (h-\lceil h/2 \rceil+1)h!\\ &= \sum_{h=0}^{2k} (\lfloor h/2 \rfloor+1)h!\\ \end{array} $
At this point, you might be able to use the fact that $\sum_{h=0}^k h h! =\sum_{h=0}^k (h+1-1) h! =\sum_{h=0}^k ((h+1)!-h!) =(k+1)!-1 $.
Another possibility is to see how $f(k)$ grows.
$\begin{array}\\ f(k+1) &=\sum_{r=0}^{k+1} \sum_{h=0}^{r} (h+r)!\\ &=\sum_{r=0}^{k} \sum_{h=0}^{r} (h+r)!+\sum_{h=0}^{k+1} (h+k+1)!\\ &=f(k)+\sum_{h=0}^{k+1} (h+k+1)!\\ &=f(k)+\sum_{h=k+1}^{2k+2} h!\\ \end{array} $
At either of these points, I don't see how to go further.
Note: I was unsure about the reversal of summations, so here is how I checked:
r=0 -> 0!
k=0 -> 1
r=1 -> 1!, 2!
k=1 -> 1
r=2 -> 2!, 3!, 4!
k=2 -> 2
r=3 -> 3!, 4!, 5!, 6!
k=3 -> 2