In one of my old (Dutch) algebra/analysis problem books, I came across the following "cute" double sum identity:
$$ \sum_{k=1}^n\sum_{h=1}^k \frac{h^2-3h+1}{h!} = - n - 1 + \frac{1}{n!}, $$
which the reader was asked to prove true for all $n\in\mathbb{N}$. The proof is relatively straightforward, with induction on $n$. (Although I'd love to see a more imaginative proof!)
My question is, can we somehow generalise this identity? Because this is a rather non-obvious result, I have a hunch that the textbook writer derived it from some more general theorem. But I have no idea what to look for.
Any insights will be warmly appreciated!
Let us consider $$S_n=\sum_{k=1}^n\sum_{h=1}^k \frac{ah^2+bh+c}{h!} $$ For the inner sum, we (say a CAS) gives (after a bunch of simplications) $$\sum_{h=1}^k \frac{ah^2+bh+c}{h!}=-\frac{-e \Gamma (k+1,1) (2 a+b+c)+a (k+2)+b}{\Gamma (k+1)}-c $$ and nothing more can be obtained.
However, if $b=-(2a+c)$, the term containing the incomplete gamma function disappears and we are left with $$S_n=\sum_{k=1}^n\sum_{h=1}^k \frac{ah^2-(2a+c)h+c}{h!}=\sum_{k=1}^n \frac{-a k-c \Gamma (k+1)+c}{\Gamma (k+1)}$$ and another bunch of simplications leads to $$S_n=\frac{e (c-a) \Gamma (n+1,1)+a-c \Gamma (n+2)}{\Gamma (n+1)}=\frac a {n!}-c(n+1)+\frac{e (c-a)}{n!} \Gamma (n+1,1)$$